Question

A 2.0 kgbreadbox on a frictionless incline of angle $\mathit{\theta }\mathbf{=}\mathbf{40}\mathbf{°}$is connected, by a cord that runs over a pulley, to a light spring of spring constantk=120N/m, as shown in Figure. The box isreleased from rest when the spring is unstretched. Assume that the pulley is mass less and frictionless. (a) What is the speed of the box when it has moved 10 cmdown the incline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude (d) direction (Up or down the incline) of the box’s acceleration at the instant the box momentarily stops?

Short answer

a) The speed of the box is $v=0.81\mathrm{m}/\mathrm{s}$.

b) The distance covered by the box along the inclined plane is x=0.21 m .

c) The magnitude of the acceleration is $a=6.3\mathrm{m}/{\mathrm{s}}^2$.

d) The direction of the acceleration is uphill.

Step-by-step solution

Step 1: Given

i) The mass of the bread box is,$m=2.0\mathrm{kg}$

ii) The inclination angle of the box is, $\theta =40°$

iii) The spring constant of the spring is,k =120 N/m

iv) The distance covered by the block is,x =10cm=0.10 m

Determining the concept

Use the concept of the energy conservation law and elastic potential energy of the spring. By using Hooke’s law, find the restoring force. To find the height, use trigonometry.According to the law of energy conservation, energy can neither be created, nor be destroyed.

Formulae are as follow:

F =-kx

$$\begin{gathered} U\left(x\right)=\frac{1}{2}k{x}^2 \\ K=\frac{1}{2}m{v}^2 \\ {F}_{net}=ma \end{gathered}$$

Where, Kis kinetic energy,U(x) is potential energy, m is mass, v is velocity, a is an acceleration, x is displacement, k is spring constant and F is force.

(a) Determining the speed of the box

Initially, the box is placed at the height as shown in the figure. At that time, the spring is up stretched. Hence, it has only gravitational potential energy. When it moves down through a distance , the vertical distance covered by the box is $∆y$. By using trigonometry,

$$\begin{gathered} \sin \theta =\frac{∆y}{x} \\ ∆y=x\sin \theta \end{gathered}$$

When the box moves through a distance x down, its gravitational potential energy is converted to kinetic energy and elastic potential energy.

According to the energy conservation law,

$$\begin{gathered} mg\Delta y=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2 \\ mg(x\sin \theta )=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2 \\ \frac{1}{2}m{v}^2=mg\left(xsin\theta \right)-\frac{1}{2}k{x}^2 \\ {v}^2=\frac{2mg(x\sin \theta )-k{x}^2}{m} \\ v=\sqrt{\frac{2mg\left(x\sin \theta \right)-k{x}^2}{m}} \\ v=\sqrt{\frac{2\times 2.0\mathrm{kg}\times 9.8{\displaystyle \frac{\mathrm{m}}{{\mathrm{s}}^2}}\times 0.10\mathrm{m}\times \sin 40-120\mathrm{N}/\mathrm{m}\times {\left(0.10\mathrm{m}\right)}^2}{2.0\mathrm{kg}}} \\ v=0.81\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the box is v=0.81 m/s.

(b) Determining the distance covered by the box along the inclined plane

The box is released along the inclined plane. Hence, the gravitational potential energy is converted to elastic potential energy. Consider, the distance covered by the box during this motion as d . According to the energy conservation law,

$$\begin{gathered} mg\Delta y=\frac{1}{2}k{d}^2 \\ mg(dsin\theta )=\frac{1}{2}k{d}^2 \\ mg(\sin \theta )=\frac{1}{2}kd \\ d=\frac{2mg(\sin \theta )}{k} \\ d=\frac{2\times 2.0kg\times 9.8{\displaystyle \frac{m}{s^2}}\times \sin 40}{120N/m} \\ d=0.21m \end{gathered}$$

Hence, the distance covered by the box along the inclined plane is x = 0.21 m.

(c) Determining the magnitude of the acceleration

The motion of the box is downhill. Hence, the restoring force acts on the box in the uphill direction. According to the figure, the component of the gravitational force is acting in the downward direction along the inclined plane. Hence, according to Newton’s second law, the net force acting on the box is,

$F_{net}=ma$

$F-mg\sin \theta =ma$ (i)

According to the Hooke’s law,

F =kx

Here,x =d

Equation (i) becomes as,

$$\begin{gathered} kd-mg\sin \theta =ma \\ a=\frac{kd-mg\sin \theta }{m} \\ a=\frac{120{\displaystyle \frac{N}{m}}\times 0.21m-2.0kg\times 9.8{\displaystyle \frac{m}{s^2}}\times \sin 40}{2.0kg} \\ a=6.3\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the magnitude of the acceleration is $a=6.3\mathrm{m}/{\mathrm{s}}^2$.

(d) Determining the direction of the acceleration

The direction of the net force is uphill. Hence, according to Newton’s second law, the acceleration of the box is uphill.

Hence, the direction of the acceleration is uphill.

The speed of the box and the distance covered by the box can be found by using the concept of conservation of energy and elastic potential energy. The acceleration of the box can be found by using Newton’s second law