Question

Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. What is the friction force of block A on block B if the block B has a mass of 24 kg and is accelerating at 0.50 m/s2 to the right relative to the block A?

A. 99 N

B. 12 N

C. 63 N

D. 75 N

Answer 1

Answer:63

Explanation: gradpoint please

thank me

Answer 2

The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.

Step-by-step explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

`F_(Anet) = m_A\cdot a_A = F - F_(fr)\\F_(Bnet) = m_B\cdot a_B = F_(fr)`

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

`a_B = a_A-0.5 (m)/(s^2)`

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

`30kg\cdot a_A = 75N - F_(fr)\\24kg\cdot a_B = F_(fr)\\a_B= a_A-0.5 (m)/(s^2)\\\implies \\a_A=(87)/(54)(m)/(s^2),\,\,\,a_B=(10)/(9)(m)/(s^2)\\F_(fr) = 24kg \cdot (10)/(9)(m)/(s^2)=(80)/(3)kg(m)/(s^2)\approx 26.7N`

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.