Question

A 17.5 kg block is dragged over a rough, horizontal surface by a constant force of 167 N acting at an angle of 30◦ above the horizontal. The block is displaced 23.9 m, and the coefficient of kinetic friction is 0.136. The acceleration of gravity is 9.8 m/s 2 . 17.5 kg µ = 0.136 167 N 30 ◦ If the block was originally at rest, determine its final speed. Answer in units of m/s.

Answer 1

Answer:

The final velocity of the block is 68.85m/s.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

$v_{f}^{2} = v_{i}^{2} + 2ad$

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$  (1)

But it is necessary to know the acceleration. For a better procedure it will be listed the knowns and unknowns of the problem:

Knowns:

F = 167 N

$\theta$ = 30°

m = 1.75 Kg

d = 23.9 m

$\mu$ = 0.136

Unknowns:

$F_{r}$ = ?

a = ?

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$  (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

Force in the x axis:

$F_{x} = F + W_{x} - F_{r}$  (3)

Forces in the y axis:

$F_{y} = N - W_{y}$ (4)

Solving for the forces in the x axis:

$F_{x} = F + W_{x} - F_{r}$

Notice that is necessary to found $F_{r}$:

$F_{r} = \mu N$  (5)

The normal force can be obtained from equation (4)

$N - W_{y} = 0$

$N = W_{y}$

The component of the weight in the y axis can be gotten by means of trigonometry:

$\frac{Adjacent}{Hypotenuse} = cos \theta$

$\frac{W_{y}}{W} = cos \theta$

$W_{y}= W cos \theta$

Remember that the weight is defined as:

$W = mg$

$W_{y}= mgcos \theta$

$N = mg cos \theta$

$F_{r} = \mu mgcos \theta$

$F_{r} = (0.136)(1.75Kg)(9.8m/s^{2})(cos30)$

$F_{r} = 2.01N$

The component of the weight in the x axis can be gotten by means of trigonometry:

$\frac{Opposite}{Hypotenuse} = sen \theta$

$\frac{W_{x}}{W} = sen \theta$

$W_{x} = W sen \theta$

$W_{x} = mgsen \theta$

$W_{x} = (1.75Kg)(9.8m/s^2)(sen30)$

$W_{x} = 8.57N$

Then, replacing $W_{x}$ and $F_{r}$ in equation (3) it is gotten:

$F_{x} = 167N + 8.57N - 2.01N$

$F_{x} = 173.56N$

Solving for the forces in the y axis:

$F_{y} = N - W_{y}$

$F_{y} = mgcos \theta - mgcos \theta$

$F_{y} = 0$

Replacing the values of $F_{x}$ and $F_{x}$ in equation (2) it is gotten:

$F_{x} + 0 = ma$

$F_{x} = ma$

$a = \frac{F_{x}}{m}$

$a = \frac{173.56N}{1.75Kg}$

$a = \frac{173.56Kg.m/s^{2}}{1.75Kg}$

$a = 99.17m/s^{2}$

Now that the acceleration is known, equation (1) can be used:

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$

However, since the block was originally at rest its initial velocity will be zero ($v_{i} = 0$).

$v_{f} = \sqrt{2ad}$

$v_{f} = \sqrt{2(99.17m/s^{2})(23.9m)}$

$v_{f} = \sqrt{2(99.17m/s^{2})(23.9m)}$

$v_{f} = 68.85m/s$

Hence, the final velocity of the block is 68.85m/s.

Answer 2

Answer:

$v = 19.03 m/s$

Explanation:

As we know that the block is dragged over the surface by external force at an angle of 30 degree above the horizontal

Now we will have

$F_n + F sin30 = mg$

$F_n = mg - F sin30$

$F_n = 17.5(9.81) - 167 sin30$

$F_n = 88.175 N$

now we know that net horizontal force on the block is given as

$F_x = Fcos30 - F_f$

$F_x = 167 cos30 - \mu F_n$

$F_x = 167cos30 - 0.136(88.175)$

$F_x = 132.6 N$

now we know that

work done by net force is equal to change in kinetic energy of the block

so we have

$F_x . d = \frac{1}{2}mv^2 - 0$

$132.6 \times 23.9 = \frac{1}{2}(17.5) v^2$

$v = 19.03 m/s$