A 200 Kg box is sitting at rest on a horizontal, flat surface; the static and th

Question

A 200 Kg box is sitting at rest on a horizontal, flat surface; the static and the kinetic friction coefficients between the box and the surface are mu_s = 0.26 and = 0.24 respectively, (use g = 10 m/s^2) Calculate the minimum force that, when applied horizontally (i.e. parallel to the ground) to the box will make the box move. Once the box is moving, the horizontally applied force is set to 600 N in the direction of the motion of the box. Calculate the magnitude of the acceleration of the box. When the box is moving at a speed of 4 m/s, the horizontal force is removed and the box slows down under the action of the kinetic friction only. Calculate the time it takes the box to come to a full stop.

Explanation / Answer

i)The minimum force is:

F(min) = mu(s) mg

F(min) = 0.26 x 200 x 9.8 = 509.6 N

Hence, F(min) = 509.6 N

ii)F = 600 N

When the box started moving, the kinetic frcitional force will act against the direction of motion. So

Fnet = F - Ff

200 x a = 600 - 0.24 x 200 x 9.78

a = 129.6/200 = 0.65 m/s^2

Hence, a = 0.65 m/s^2

iii)when no force is acting, the acceleration will be

ma = mu(k) mg => a = mu(k)g = 0.24 x 9.8 = 2.35 m/s^2

Its actually retardation.

time required to decelerate from 4m/s to 0 will be:

t = (0 - 4)/-2.35 = 1.702 s

Hence, t = 1.702 s

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