In Figure (a), a 1.9 N force is applied to a 5.1 kg block at a downward angle th

Question

In Figure (a), a 1.9 N force is applied to a 5.1 kg block at a downward angle theta = 57degree as the block moves rightward through 1.3 m across a frictionless floor. Find the speed v_f of the block at the end of that distance if the block's initial velocity is (a) 0 and (b) 1.5 m/s to the right. (c) The situation in Figure (b) is similar in that the block is initially moving at 1.5 m/s to the right, but now the 1.9 N force is directed downward to the left. Find the speed v_f of the block at the end of the 1.3 m distance. Number Units Number Units Number Units By accessing the Question Assistance, you will learn while you earn points based on the point potential policy set by your instructor.

Explanation / Answer

Part A.) Here the block is being acted upon by a force F = 1.9 N at an angle of 57 degrees with the horizontal. Now, we know that the work done by the force on the block would be given as the dot product of the force and the displacement.

That is, W = F.S = FS Cos = 1.9 x 1.3 x Cos57 = 1.34526 J

Now, we will use the principle of conservation of energy to get the kinetic enegry the block finally attains,

Hence, KE (Initial) + W = KE(Final)

That is, 0 + 1.34526 = 0.5 x 5.1 x v2

or, V = 0.7263 m/s is the required velocity

Part B.) Here again, we have the ent work done on the block as: 1.34526 J

Also, it will have an initial kinetic energy. So, we get:

0.5 x 5.1 x 2.25 + 1.34526 = 0.5 x 5.1 x v2

or, V = 1.6666 m/s is the required velocity

Part C.) Here the direction of the force changes, hence the net work done will be negative.

So we get: 0.5 x 5.1 x 2.25 - 1.34526 = 0.5 x 5.1 x v2

or, v = 1.3124 m/s is the required velocity

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