1
$\begingroup$

Assume a block with mass $m_b$ is on top of a cart with mass $m_c$. The coefficient of kinetic friction between the block and the cart is $μ_k$, the coefficient of static friction between the block and the cart is $μ_s$, and the wheels of the cart are of negligible mass and rotate with negligible friction.

As the block and cart move to the right, which has a larger magnitude? The work done by the force of kinetic friction acting on the block, or the work done by the force of kinetic friction acting on the cart?

At first, I thought the work done would be the same because the two forces are a Newton's force pair, and they act over the same distance on the top of the cart. However, I checked the answer key and realized that the work done by kinetic friction on the cart is greater because the cart has a greater acceleration and, therefore, a greater displacement, which leads to more work done because the equation for work is $W = Fd$.

Why is this? I know objectively from the point of view of an object at rest the cart seems to travel a greater distance, but from the point of view of the block they both travel the same distance away from each other, so shouldn't the work done be the same?

Improve this question
$\endgroup$
5
  • $\begingroup$ Energy and work aren't reference frame independent, so in order to say which object has more work done to it, we need to use the same reference frame for both. Also, from an object's OWN reference frame, it's always completely still, meaning no work is being done to it, so it's not correct to say that, from the block's reference frame, work is being done to it. Most likely, the question is assuming a reference frame of an observer standing on the ground, watching the motion of the cart and block. Generally, in classical mechanics problems like this, unless otherwise specified, the assumed $\endgroup$
    – Mikayla Eckel Cifrese
    Commented May 9, 2023 at 21:09
  • 1
    $\begingroup$ Hello, thank you for your response! Sorry if I worded my question incorrectly, but what I meant to convey was that I was having trouble understanding the concept of work in this problem rather than frames of reference. I keep getting problems like this that test my conceptual understanding of work wrong. I still don't quite understand, so could you clarify the concept or suggest a resource? Thanks, @ Mikayla Eckel Cifrese $\endgroup$
    – Aryaman Jaggi
    Commented May 9, 2023 at 21:31
  • $\begingroup$ I posted an answer explaining where you want wrong with analyzing the work done. That said, homework type questions don't tend to be very well received on here, so for future reference, the homework help subreddit might be a better fit for this type of question. $\endgroup$
    – Mikayla Eckel Cifrese
    Commented May 9, 2023 at 23:45
  • $\begingroup$ Is the block moving relative to the cart? $\endgroup$
    – nasu
    Commented May 10, 2023 at 1:46
  • $\begingroup$ Yes, the cart is being pulled to the right by a horizontal force, and the block which is on top of the cart is sliding due to kinetic friction, so the cart is accelerating faster than the block. $\endgroup$
    – Aryaman Jaggi
    Commented May 11, 2023 at 2:06

1 Answer 1

Reset to default
0
$\begingroup$

Remember, work is force times distance, and force is proportional to mass, and frictional force is just the product of the normal force (which in this case is just gravity) and the coefficient of friction. The normal force on the block will be its mass times $g$, but since the block is on top of the cart, the normal force on the cart will be $g$ times the sum of the mass of block and the mass of the cart. Hence the force on the cart will clearly be larger than the force on the block. And since, as you already correctly inferred, the distance travelled by both objects will be the same (at least assuming the block doesn't get pushed off the cart (but even if it did, it wouldn't be friction causing that)), the work done on the cart must be larger.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.