Question

A student, crazed by final exams, uses a force of magnitude 80Nand angle$\mathit{\theta }\mathbf{=}\mathbf{70}\mathbf{°}$to push a 5.0 kg block across the ceiling of his room (Fig. 6-52). If the coefficient of kinetic friction between the block and the ceiling is 0.40, what is the magnitude of the block’s acceleration?

Short answer

$a=3.4\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

Given data

• The push force,$P=80\mathrm{N}$.
• The angle,$\theta =70°$.
• The mass of the block,$m=5\mathrm{kg}$.
• The coefficient of friction, ${\mu }_k=0.4$.

To understand the concept

The problem deals with Newton’s second law of motion, which states that the acceleration of an object can be calculated with the help of the net force acting upon the object and the mass of the object. Start by calculating the normal force acting on the box. After calculating the normal force, we will use Newton’s second law of motion to calculate the acceleration.

Calculate the magnitude of the block’s acceleration

As the acceleration is in the vertical direction,

$F_N=P\sin \theta -mg$

The frictional force acting on toward left is given by:

$f_k={\mu }_k\left(P\sin \theta -mg\right)$

Choosing +x rightward, Newton’s second law of motions gives us:

$$\begin{gathered} P\cos \theta -f_k=ma \\ a=\frac{P\cos \theta -{\mu }_k\left(P\sin \theta -mg\right)}{m} \end{gathered}$$

Substitute the values in the above expression, and we get,

$\mathrm{a}=\frac{\left(80\mathrm{N}\right)\cos \left(70°\right)-0.4\times \left(\left(80\mathrm{N}\right)\sin \left(70°\right)-\left(5\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\right)}{5\mathrm{kg}}$

$\mathrm{a}=3.4\mathrm{m}/{\mathrm{s}}^2$

Thus, the magnitude of the acceleration of the block is $3.4\mathrm{m}/{\mathrm{s}}^2$.