a 1.8 N force is applied to a 4.2 kg block at a downward angle = 35° as the bloc

Question

a 1.8 N force is applied to a 4.2 kg block at a downward angle = 35° as the block moves rightward through 1.4 m across a frictionless floor. Find the speed vf of the block at the end of that distance if the block's initial velocity is (a) 0 and (b) 1.4 m/s to the right. (c) The situation in Figure 7-47b is similar in that the block is initially moving at 1.4 m/s to the right, but now the 1.8 N force is directed downward to the left. Find the speed vf of the block at the end of the 1.4 m distance.

Explanation / Answer

Horizontal force = 1.8cos35 = 1.474N

a = F/m = 1.474/4.2 = 0.35m/s^2

(a) v = (2*a*s) = (2*0.35*1.4) = 0.9899m/s

(b) v = (2as+u^2) = (2*0.35*1.4 + 1.4^2) = 1.714m/s

(c) v = (-2*0.35*1.4 + 1.4^2) = 0.9899m/s

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