SFx = 15N - 7Ncos25?= 8.66N.
We find the horizontal acceleration of the block from
ax = SFx /m = 8.66N/5kg
= 1.73m/s2.(to the right)
We find the distance from
x = x0 + v0t + 1/2at2 = 0 + 0 + 1/2(1.73m/s2(10s)2 = 86.5m.(to the right)
QUIZ 3
1. Two forces exert on a 5-kg block originally at rest as shown in Figiure. How far will the block move horizontally in 10s?.
Solution.
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We find the horizontal net force exerted on the block from
SFx = 15N - 7Ncos25?= 8.66N. We find the horizontal acceleration of the block from ax = SFx /m = 8.66N/5kg = 1.73m/s2.(to the right) We find the distance from x = x0 + v0t + 1/2at2 = 0 + 0 + 1/2(1.73m/s2(10s)2 = 86.5m.(to the right) |