Question

1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal. The coefficient of friction between the crate and the floor is 0.12.
a) Calculate the net force and acceleration of the crate.

b) If the crate was moving at 1.0 m/s when it was pulled, what would be its velocity after pulling it for 5.0 s?

Answer 1

Answer:

(a) The net force is 80.394 N

    The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, $F_y = Fsin \theta = 250sin45 = 176.78 \ N$

Applied force in x-direction, $F_x = Fcos \theta = 250cos45 = 176.78 \ N$

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

$F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N$

Apply Newton's second law of  motion;

F = ma

$a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2$

(b) the velocity of the crate after 5.0 s

$F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s$