Block in the figure weighs 1.40 and block weighs 3.80 . The coefficient of kinet

Question

Block in the figure weighs 1.40 and block weighs 3.80 . The coefficient of kinetic friction between all surfaces is 0.400. (Figure 1) Part A Find the magnitude of the horizontal force necessary to drag block to the left at a constant speed of 7.00 if rests on and moves with it (figure a). = SubmitMy AnswersGive Up Part B Find the magnitude of the horizontal force necessary to drag block to the left at a constant speed of 7.00 if is held at rest by a string (figure b). = SubmitMy AnswersGive Up Part C In part (A), what is the friction force on block ? = SubmitMy AnswersGive Up Provide FeedbackContinue of 1

Explanation / Answer

Part A)
The combined mass of the blocks is 5.20
In order to make the blocks move at a constant speed, we need to apply a force that is equal to the force of friction.
We can calculate the force of friction as F =N where is the coefficient and N is the Normal force.

The normal force is mg

N = (5.20)(9.8)

The coefficient is 0.400

So the force from friction is (0.400)*(5.20)*(9.80) = 20.384

So the answer is 20.384 Newtons

Part B)

We will have to sum the force of Friction from the ground on block B and from block A on block B.

From the ground:

F = (0.400)(3.80)(9.80) = 14.896

From block A:

F= (0.400)(1.40)(9.80) = 5.488

When we sum, we get the same answer, 20.384.

So the answer is 20.384

Part C)

I can't see if you ar asking about block A or block B

We calculated the force of friction on block B, it is 20.384

Since Block A is not moving relative to block B, the force from friction must be 0

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