A student decides to move a box of books into her dormitory room by pulling on a

Question

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 79.2 N at an angle of 27.0° above the horizontal. The box has a mass of 20.0 kg, and the coefficient of kinetic friction between box and floor is 0.300. (Indicate the direction with the sign of your answer.)

(a) Find the acceleration of the box. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)

(b) The student now starts moving the box up a 10.0° incline, keeping her 79.2 N force directed at 27.0° above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box? (Assume that the +x-axis is along the incline and the +y-axis is upwards and perpendicular to the incline.)

Explanation / Answer

use the sum of Force along x direction = 0

i,e F cos 270 -Ffric

= 79.2 * cos 270 - 0.3*(Normal reaction)                 ; (Assume Normal reaction=N)

= 70.57 - 0.3 N

now using vertical sum of forces = N + 79.2 sin 270 = mg

N = 20*9.8 - 35.95 = 160.1

so now by substituing Net force = 70.57- (0.3 * 160.1)

force N = 22.54

accleration a = F/m = 22.54/20 = 1.127 m/s^2

Direction = zero degree counterclockwise

-------------------------------------

B.
normal force. 20*9.8*cos100= 193 N

Friction Force = friction coefficient * normal force

Ff = 0.3*193 = 57.9 N.

The force parallel to the incline, going up is still 79.2 cos27.

Fnet = m*a

79.2 cos27 - 57.9 = 20 * a

a = 0.6335 m/s^2

Direction: Along the incline , Zero degree counterclockwise

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