Question

A 1.00 -kg block is pushed up and down a rough plank of length \(L=2.00 \mathrm{~m},\) inclined at \(30.0^{\circ}\) above the horizontal. From the bottom, it is pushed a distance \(L / 2\) up the plank, then pushed back down a distance \(L / 4,\) and finally pushed back up the plank until it reaches the top end. If the coefficient of kinetic friction between the block and plank is \(0.300,\) determine the work done by the block against friction.

Short answer

Answer: The total work done by the block against friction is 2.54 J.

Step-by-step solution

Calculate the frictional force

The frictional force \(F_f\) acting on the block is given by the formula \(F_f = μN\), where \(μ\) is the coefficient of friction and \(N\) is the normal force acting on the block. In this case, μ is given as \(0.300\). Now, to calculate the normal force, we need to consider the weight of the block, which is \(mg\) (where m is the mass and g is the gravitational acceleration). Since the plank is inclined at an angle of \(30.0^{\circ}\), the normal force component will be \(mg\cos(30.0^{\circ})\). Then, we have: \(F_f = μmg\cos(30.0^{\circ}) = 0.300 \times 1.00~\text{kg} \times 9.81~\text{m/s^2} \times \cos(30.0^{\circ})\) Calculating this, we find that the frictional force is approximately \(F_f = 2.54~\text{N}\).

Calculate the work done against friction in each part of the motion

We have 3 parts of motion to consider: moving up L/2, moving down L/4, and moving up L/4. In each part, the work is given by the formula \(W = F \cdot d \cdot \cos(\theta)\). Since the force and distance are always either in the same direction (when moving up) or opposite directions (when moving down), the angle θ will be either \(0^{\circ}\) or \(180^{\circ}\). This simplifies the cosine term to either 1 (when moving up) or -1 (when moving down). Part 1 - Moving up L/2: \(W_{1} = F_f \cdot \frac{L}{2} \cdot \cos(0^{\circ}) = 2.54~\text{N} \cdot 1.00~\text{m} \cdot 1 = 2.54~\text{J}\) Part 2 - Moving down L/4: \(W_{2} = F_f \cdot \frac{L}{4} \cdot \cos(180^{\circ}) = 2.54~\text{N} \cdot 0.50~\text{m} \cdot (-1) = -1.27~\text{J}\) Part 3 - Moving up L/4: \(W_{3} = F_f \cdot \frac{L}{4} \cdot \cos(0^{\circ}) = 2.54~\text{N} \cdot 0.50~\text{m} \cdot 1 = 1.27~\text{J}\)

Sum the work done in each part to find the total work against friction

Now, we just need to sum up the work done against friction in each part of the motion to find the total work: \(W_{\text{total}} = W_{1} + W_{2} + W_{3} = 2.54~\text{J} + (-1.27~\text{J}) + 1.27~\text{J} = 2.54~\text{J}\) The total work done by the block against friction is \(2.54~\text{J}\).

Key concepts

Kinetic Friction

Kinetic friction is the force that opposes the motion of an object sliding across a surface. In this scenario, a block moves over a rough inclined plank. The kinetic friction comes into play each time the block moves.
This frictional force depends on two main factors:

• The coefficient of kinetic friction (\(μ\))—which is a measure of how rough the surfaces are.
• The normal force (\(N\)) perpendicular to the surface.

The irksome part about kinetic friction is that it always acts opposite to the direction of motion, making it much harder to move objects up an inclined plane. Understanding this relationship helps when calculating the work required to move an object over a given distance.

Inclined Plane

An inclined plane is a flat supporting surface tilted at an angle, other than a right angle, against a horizontal surface. The inclination makes it easier or necessary to work against gravity. In this problem:

• The plank is inclined at an angle of (\(30^{\circ}\)).
• This inclination affects both the normal force and the frictional force.

Understanding the effects of an inclined plane includes knowing that the weight of the object can be split into two components: parallel and perpendicular to the plane. The parallel component affects how the object slides down due to gravity, while the perpendicular component determines the normal force.

Frictional Force

The frictional force (\(F_f\)) is a resistant force evident when an object moves or attempts to move across a surface. Its magnitude is determined by:

• The coefficient of kinetic friction (\(μ\)).
• The normal force (\(N\)).

For an object on an inclined plane, the normal force is less than the object's weight due to the angle of inclination. Here’s how it’s calculated:
\[ F_f = μmg\cos(\theta) \]
Where \( \theta \) is the angle of the incline. In our case, \(F_f\) is affected by the inertia of the block and the downward pull of gravity. Understanding this helps us calculate the work done to move against this resistive force.

Normal Force

Normal force is the contact force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface.
When an object is on a flat surface, the normal force is equal to its weight. However, for an inclined plane:

• The normal force is equal to
  \(mg\cos(\theta)\).

In this exercise, the angle makes the normal force less than the object's weight. The normal force plays a crucial role in determining the frictional force (since friction depends on it). Understanding normal force helps in calculating how much energy is required to overcome friction when moving objects along an inclined plane.