A worker drags a crate across a factory floor by pulling on a rope tied to the crate. The worker exerts a force of magnitude f = 450Non the rope, which is inclined at an upward angle$\theta ={38}^{°}$ to the horizontal, and the floor exerts a horizontal force of magnitude f = 125 Nthat opposes the motion. Calculate the magnitude of the acceleration of the crate if (a) its mass is 310kg and (b) its weight is 310 N.

It is given that,

$$\begin{gathered} F_1=450\mathrm{N} \\ {F}_2=125\mathrm{N} \end{gathered}$$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Using Newton’s second law of motion, write the acceleration of the given object and solve it for different conditions given. Component of the force can be found using vector resolution method along the required directions.

Formula is as follow:

The net force is given by,

$F_{\mathrm{net}}=\sum _{}^{}Ma$

By applying Newton’s second law:

$F_1\cos \left(\theta \right)-F_2=ma$

$a=\frac{F_1\cos \left(\theta \right)-F_2}{m}$

When $F_1\cos \left(\theta \right)-F_2\hspace{0.17em}=ma$

$a=\frac{450\cos \left(38\right)-125}{310}$

$a=0.74\mathrm{m}/{\mathrm{s}}^2$

Hence, the magnitude of the acceleration of the crate when m= 310 kg is $a=0.74\mathrm{m}/{\mathrm{s}}^2$

When W = 310 N

Hence, mass

$\left(\mathrm{m}\right)=\frac{310}{9.8}$

$=31.6\mathrm{kg}$

Now,

$a=\frac{450\cos \left(38\right)-125}{31.6}$

$a=7.3\mathrm{m}/{\mathrm{s}}^2$

Hence, magnitude of the acceleration of the crate when W = 310N is$a=7.3\mathrm{m}/{\mathrm{s}}^2$