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 168 N at an angle of 26.0° above the horizontal. The box has a mass of 21.0 kg, and the coefficient of kinetic friction between box and floor is 0.300.

(a) Find the acceleration of the box.
(b) The student now starts moving the box up a 10.0° incline, keeping her 168 N force directed at 26.0° above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box?

Explanation / Answer

use the summ of Force along x direction = 0

i,e F cos 26 -Ff

= 168 * cos 26 - 0.3 N

= 151 -0.3 N

now using vertical smm of forces = N + 168 sin 26 = mg

N = 132.14

so now by substituing Net force = 151-(0.3 * 132.14)

force N = 111.358 N

accleration a = F/m = 111.358/21 = 5.32 m/s^2

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B.
normal force. 21*9.8*cos10= 202.67 N

Friction = your coefficient * normal force

Ff = 0.3*202.67 = 60.801 N.

The force parallel to the incline, going up is still 168 cos26.

Fnet = m*a

168 cos26 - 60.88 = 21 * a

a = 4.291 m/s^2

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