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## Homework Statement

A 45.0 kg box is pulled with a force of 205 N

by a rope held at an angle of 46.5° to the

horizontal. The velocity of the box increases

from 1.00 m/s to 1.50 m/s in 2.50 s. Calculate

a) Net force acting horizontally on the box

b) Frictional force on the box

c) Horizontal component of the applied force

d) Co-efficient of kinetic friction between the box and the floor

## Homework Equations

Kinematics equations

[mu][Fn] = frictional force

## The Attempt at a Solution

a) I found acceleration by: change in velocity = 0.5m/s, change in time = 2.50s

v

---

t

= 0.2m/s^2

Ok I found the summation of the X and Y forces:

b) Summation of X forces: [(205)(cos46.5) - frictionalF = (45)(0.2)]

frictionalF = 132.1N [<---]

Summation of Y forces: [Fn + (205)(sin46.5) - (45.0)(0.2) = 0]

normalF = 292.74N

c) Now, I found the horizontal component of the applied force to be:

[205][cos46.5] = 141.1N, but the answer at the back says 293N. Have I done something wrong?

d) I am not sure how I am going to get the coefficient.