Question

A $3.5kg$block is pushed along a horizontal floor by a force of magnitude $\mathbf{15}\mathbf{}\mathit{N}$at an angle with the horizontal (Fig. 6-19). The coefficient of kinetic friction between the block and the floor is $\mathbf{0}\mathbf{.}\mathbf{25}$. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block’s acceleration.

Short answer

(a) The magnitudes of the frictional force on the block from the floor is 11 N.

(b) The magnitudes of the block’s acceleration on the block from the floor is $0.14\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Step-by-step solution

Given

Mass of the block, $m=3.5kg$

Force on the block, $F=15\text{N}$

Angle, $\theta ={40}^0$

Coefficient of kinetic friction,${\mu }_k=0.25$

Determining the concept

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. Use the Newton's 2nd law of motion along vertical and horizontal direction. Also, using the free body diagram, the magnitude of frictional force and acceleration can be found,

Write the formula for the net force as:

$F_{net}=\sum ma$

Here, F is the net force, mis mass and ais an acceleration.

(a) Determining themagnitudes of the frictional force on the block from the floor

By using Newton’s 2nd law along y direction.

Block is not moving along y.So,the acceleration is 0.

$$\begin{gathered} \sum _{}^{}{F}_y=0 \\ N-mg-15\sin \left(40\right)=0 \\ N=mg+15\sin \left(40\right) \\ =44\text{N} \end{gathered}$$

Frictional force on the block from the floor is,

$f_k={\mu }_{k}N$

Substitute the values and solve as:

$\begin{array}{rcl}{f}_k& =& \left(0.25\right)\left(44\right)\\ & =& 11\text{N}\\ & & \end{array}$

Hence, the magnitudes of the frictional force on the block from the floor is 11 N.

(b) Determine theblock’s acceleration

According to the Newton’s 2nd law along horizontal direction i.e. x is as the block moving along x. Let, the acceleration of the block is a.

$\sum _{}^{}{F}_x=m{a}_x$

Substitute the values and solve for acceleration as:

$\begin{array}{rcl}a& =& \frac{15\cos \left(40\right)-f_k}{m}\\ & =& 0.14\frac{\text{m}}{{\text{s}}^{\text{2}}}\\ & & \end{array}$

Hence, the magnitudes of the block’s acceleration on the block from the floor is $0.14\frac{\text{m}}{{\text{s}}^{\text{2}}}$.