Question

A(n) 930 N crate is being pushed across a level floor by a force of 400 N at an angle of 20◦ above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.20. The acceleration of gravity is 9.81 m/s 2 . What is the magnitude of the acceleration of the box? Answer in units of m/s 2

Answer 1

Answer:

The magnitude of the acceleration of the box is $2.291\,\frac{m}{s^{2}}$.

Explanation:

The free body diagram of the crate is included as attachment, whose equations of equilibrium are described below:

$\Sigma F_{x} = P\cdot \cos 20^{\circ} - \mu_{k}\cdot N = \left(\frac{W}{g}\right)\cdot a$

$\Sigma F_{y} = P\cdot \sin 20^{\circ} + N - W = 0$

From second equation of equilibrium we find an expression for the normal force and find the respective value:

$N = W - P \cdot \sin 20^{\circ}$

$N = 930\,N - 400\cdot \sin 20^{\circ}\,N$

$N = 793.192\,N$

Lastly, the acceleration experimented by the crate during pushing is cleared in the first equation of equilibrium and consequently calculated:

$a = \frac{P\cdot \cos 20^{\circ}-\mu_{k}\cdot N}{\frac{W}{g} }$

$a = \frac{400\cdot \cos 20^{\circ}\,N-0.20\cdot (793.192\,N)}{\frac{930\,N}{9.807\,\frac{m}{s^{2}} } }$

$a = 2.291\,\frac{m}{s^{2}}$

The magnitude of the acceleration of the box is $2.291\,\frac{m}{s^{2}}$.