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A 0.800-kg ball is tied to the end of a string 1.60 m long and swung in a vertical circle. (a) During one complete circle, starting anywhere, calculate the total work done on the ball by (i) the tension in the string and (ii) gravity. (b) Repeat part (a) for motion along the semicircle from the lowest to the highest point on the path.

Short Answer

Expert verified
(a) Zero for both; (b) Zero by tension, 25.088 J by gravity.

Step by step solution

01

Understand the Problem

We have a 0.800-kg ball tied to a 1.60 m long string, swung in a vertical circle. We need to calculate the work done by the tension in the string and the gravitational force in two scenarios: a complete circle (part a) and a semicircle from the lowest to the highest point (part b).
02

Calculate Work Done by Tension in Complete Circle

Work done by tension is calculated with the formula W=Fcos(θ)d where F is the force, θ is the angle between force and displacement, and d is displacement. Since tension is perpendicular to displacement through the complete circle, θ=90 and cos(90)=0. Therefore, work done, W=0.
03

Calculate Work Done by Gravity in Complete Circle

The gravitational force does work, W=mgh, where h is the change in height. In a complete circle, the ball returns to the starting point, so the change in height is zero. Consequently, the work done by gravity is zero, W=0.
04

Calculate Work Done by Tension for Semicircle

In a vertical motion from the lowest to the highest point of the semicircle, the tension remains perpendicular to the instantaneous displacement at every point. Thus, similar to the full circle, the work done by the tension remains W=0.
05

Calculate Work Done by Gravity for Semicircle

From the lowest to the highest point, the vertical height change is twice the radius, 2r=2×1.60×sin(θ). At the top, θ=90, so the height change is 3.20 m. Work done by gravity is W=mgh=0.800×9.8×3.20. Calculate W=25.088 J.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circular Motion
Circular motion occurs when an object moves in a path in the shape of a circle. This type of motion is characterized by a constant change in direction, even if the speed of the object remains constant.
In a circular motion, the velocity of the object is always tangential to the circle's path. This means the direction of the velocity vector changes as the object moves, even if the speed (magnitude of the velocity) remains constant.
To maintain this motion and change direction continuously, a centripetal force is needed, which always points toward the center of the circle.
  • The centripetal force is not a separate force but the resultant of other forces acting on the object.
  • In the case of a ball tied to a string swung in a circle, the tension in the string contributes to this centripetal force.
Understanding circular motion is key to studying how forces interact in a rotating system and how different scenarios, like vertical circles, affect these dynamics.
Vertical Circle
Vertical circular motion involves an object moving in a circle located in a vertical plane. This adds complexity because, as the object moves, the gravitational force affects it differently at varying points.
One key factor in vertical circular motion is how gravitational force interacts with other forces like tension. Unlike horizontal motion, gravity contributes to the dynamics by accelerating or decelerating the object as it rises and falls.
When the object is at the highest point in the circle, gravity acts downward, potentially decreasing tension in the string. Conversely, when the object is at the lowest point, gravity increases tension since it acts in the same direction.
  • Gravitational force is crucial because it can lead to changing tension forces throughout the motion.
  • Understanding these dynamics enables the calculation of forces and predictability in motion outcomes, especially when calculating work done by gravity and tension.
Vertical circles provide an excellent real-world application, whether in amusement park rides or understanding planetary motion.
Work Done by Gravity
Work done by gravity plays a crucial role in many physics problems involving motion. It is the energy transferred by the gravitational force over a distance.
In the context of a ball in circular motion, the work done by gravity depends on the vertical displacement of the ball. When the ball returns to its starting height, the work done by gravity over the complete circle is zero, as there is no net change in height.
  • Formula: The work done by gravity is given by W=mgh where m is the mass, g is the acceleration due to gravity, and h is the change in height.
  • During vertical motion through a semicircle, however, there is a net vertical displacement, hence work done is not zero.
  • This calculation helps determine how potential energy converts to kinetic energy and vice-versa through motion.
Understanding work done by gravity aids in analyzing the energy changes and forces at play in cyclic motion.
Tension in Strings
Tension in strings is a vital concept when studying objects in circular motion. It is the force exerted along the string that transmits a pulling force from one end to the other.
When an object moves in a circular path, the tension helps keep the object in motion by providing the necessary centripetal force toward the center of the circle.
In our scenario with the ball on a string, tension remains perpendicular to the displacement because the ball moves along the arc of the circle. As a result, the work done by tension is zero when the object completes a full cycle or semicircle.
  • The tension force acts radially, maintaining the grip needed for circular motion.
  • Its value changes depending on the ball's position in the circle, especially impacted by the gravitational pull.
  • The perpendicular nature of tension regarding motion direction means that it does no work (i.e., transfers no energy over a distance).
Comprehending how tension works in strings forms the basis for solving complex motion and force problems in physics.

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Most popular questions from this chapter

A 5.00-kg block is moving at υ0 = 6.00 m/s along a frictionless, horizontal surface toward a spring with force constant k = 500 N/m that is attached to a wall (Fig. P6.79). The spring has negligible mass. (a) Find the maximum distance the spring will be compressed. (b) If the spring is to compress by no more than 0.150 m, what should be the maximum value of υ0?

A block of ice with mass 2.00 kg slides 1.35 m down an inclined plane that slopes downward at an angle of 36.9 below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction.

A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0 above the horizontal by a rope exerting a 72.0-N pull parallel to the ramp's surface. If the carton travels 5.20 m along the surface of the ramp, calculate the work done on it by (a) the rope, (b) gravity, and (c) the normal force of the ramp. (d) What is the net work done on the carton? (e) Suppose that the rope is angled at 50.0 above the horizontal, instead of being parallel to the ramp's surface. How much work does the rope do on the carton in this case?

We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass M, equilibrium length L0, and force constant k. The work done to stretch or compress the spring by a distance L is \(\frac{1}{2}\) kX2, where X=LL0. Consider a spring, as described above, that has one end fixed and the other end moving with speed v. Assume that the speed of points along the length of the spring varies linearly with distance l from the fixed end. Assume also that the mass M of the spring is distributed uniformly along the length of the spring. (a) Calculate the kinetic energy of the spring in terms of M and v. (Hint: Divide the spring into pieces of length dl; find the speed of each piece in terms of l, v, and L; find the mass of each piece in terms of dl, M, and L; and integrate from 0 to L. The result is not \(\frac{1}{2}\) Mv2, since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 kg and force constant 3200 N/m is compressed 2.50 cm from its unstretched length. When the trigger is pulled, the spring pushes horizontally on a 0.053-kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncompressed length (b) ignoring the mass of the spring and (c) including, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

You are applying a constant horizontal force F=(8.00N)ı^+(3.00N)ȷ^ to a crate that is sliding on a factory floor. At the instant that the velocity of the crate is υ=(3.20m/s)ı^+(2.20m/s)ȷ^, what is the instantaneous power supplied by this force?

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