Question

John was pushing a 6.8 kg box horizontally with a force of 7.0 N at a steady speed of 0.2 m/s. Then he decided to tie a rope to the box to make it possible to pull it forward. Now the box is moving at 0.6 m/s as John is pulling on it at the angle of 30.3o up of horizontal. What is the magnitude of the force of John's pull?

Answer 1

here,

mass of box , m = 6.8 kg

initial it is moving at constant speed ,

so, the force applied , F1 = friction force

7 = uk * m * g

7 = uk * 6.8 * 9.81

solving for uk

uk =0.105

when theta = 30.3 degree

let the force applied by john be F

as it is moving at constant velocity

the net force is zero

equating the forces

uk * ( m * g - F * sin(theta)) = F * cos(theta)

0.105 * ( 6.8 * 9.81 - F * sin(30.3)) = F * cos(30.3)

solving for F

F = 7.64 N

the force applied by John's pull is 7.64 N