Question Video: Determining the Force of Friction for a Sled Travelling Down a Slope | Nagwa Question Video: Determining the Force of Friction for a Sled Travelling Down a Slope | Nagwa

# Question Video: Determining the Force of Friction for a Sled Travelling Down a Slope Physics • First Year of Secondary School

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A child carries a sled to the top of an evenly sloping hill. The child puts the sled on the slope where it is just held in place by friction and carefully climbs on board. The added weight of the child is just enough to start the sled moving and it slides down the hill. The sled travels 25 m along the slope. The gravitational potential energy of the child and sled at the top of the hill is 3,500 J and at the bottom of the slope it is zero. The kinetic energy of the child and sled when they arrive at the base of the hill is 3,125 J. What average force of friction does the hillside apply to the sled during the sled’s motion?

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### Video Transcript

A child carries a sled to the top of an evenly sloping hill. The child puts the sled on the slope where it is just held in place by friction and carefully climbs on board. The added weight of the child is just enough to start the sled moving and it slides down the hill. The sled travels 25 meters along the slope. The gravitational potential energy of the child and sled at the top of the hill is 3,500 joules and at the bottom of the slope it is zero. The kinetic energy of the child and sled when they arrive at the base of the hill is 3,125 joules. What average force of friction does the hillside apply to the sled during the sled’s motion?

To answer this question, we need to think about the energy transfers that occur as the child moves down the hill. Let’s start by going through the information we’ve been given in the question.

At the top of the hill, the child and the sled together have a gravitational potential energy of 3,500 joules. At this point, the sled is stationary, so the kinetic energy must be zero. At the bottom of the hill, the child and sled have a kinetic energy of 3,125 joules but a gravitational potential energy of zero. Before we continue, let’s clear some space near the top here.

Now then, recall that the mechanical energy of an object is equal to the sum of the gravitational potential energy and the kinetic energy. At the top of the hill, the mechanical energy of the child and the sled is equal to 3,500 joules plus zero joules, which is simply equal to 3,500 joules. At the bottom of the hill, the mechanical energy is equal to zero joules plus 3,125 joules, which is just 3,125 joules. If we compare these values, we see that the mechanical energy of the child and sled is greater at the top of the hill than at the bottom.

We know that in a closed system, energy is always conserved. So how can we explain the difference between these two values? Well, we know that there’s some friction between the sled and the hill. Friction is a force that acts in the opposite direction to the motion of an object. Friction does work on the object and causes the kinetic energy of the object to decrease. This causes some energy to be dissipated. The amount of energy that is dissipated is equal to the work done by the frictional force. And this work is simply equal to the difference between the initial mechanical energy of the child and sled and the final mechanical energy. So, the work done by friction is equal to 3,500 joules minus 3,125 joules. So, the work done by the frictional force is equal to 375 joules.

Next, we need to find the average magnitude of the frictional force between the hillside and the sled. To do this, we need to recall the formula for the work done by a force. The work done 𝑊 is equal to the average force 𝐅 that is exerted on an object multiplied by the distance 𝑑 through which the object moves. To find the average frictional force, we simply need to rearrange this equation to make 𝐅 the subject. This can be done by dividing both sides of the equation by 𝑑. This leaves us with the formula 𝐅 equals 𝑊 over 𝑑.

The last step is substituting in the values of 𝑊 and 𝑑 into this equation. We have already calculated that the work done by the force 𝑊 is equal to 375 joules. The question tells us that the sled travels 25 meters along the slope, so this is our value of 𝑑. Substituting these values in, we see that the average frictional force is equal to 375 joules divided by 25 meters. This gives us an answer of 15 newtons.

So, the hillside applies an average frictional force of 15 newtons to the sled during its motion. 15 newtons is our final answer to this question.

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