Answer to Question #176962 in Physics for jowel

Question

2. A block of mass m sitting on a frictionless inclined plane is tied to a cord, and the tension in the cord is 20 N. The plane makes an angle 30° with the horizontal. (a) Find the mass of the block. (b) Find the normal force acting on the block (c) If the cord is cut, find the magnitude of the resulting acceleration of the block.

#3. The position of a 5 kg object changes according to the following relationship when a single force is applied to it 𝑥 = 2 𝑡 3 − 3 𝑡 2 + 4 (𝑡 𝑖𝑛 second 𝑎𝑛𝑑 𝑥 𝑖𝑛 𝑚𝑒𝑡𝑒𝑟) Find the work done on the object by the force from 𝑡 = 0 𝑡𝑜 𝑡 = 5 𝑠.

#5. A 5 kg block initially at rest is pulled to the right along a horizontal path having a coefficient of kinetic friction 0.33 by a constant horizontal force of 20 N. Determine the speed and acceleration of the block.

Answer 1

2) a)

$$T-mgsin\theta=0,$$ $$m=\dfrac{T}{gsin\theta}=\dfrac{20\ N}{9.8\ \dfrac{m}{s^2}\cdot sin30^{\circ}}=4.08\ kg.$$

b)

$$N-mgcos\theta=0,$$ $$N=mgcos\theta=4.08\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot cos30^{\circ}=34.63\ N.$$

c)

$$mgsin\theta=ma,$$ $$a=gsin\theta=9.8\cdot sin30^{\circ}=4.9\ \dfrac{m}{s^2}.$$

3) Let's first find the displacements of the object at $t=5\ s$ and $t=0\ s$:

$$d_1=2\cdot(0\ s)^3-3\cdot(0\ s)^2+4=4\ m,$$ $$d_2=2\cdot(5\ s)^3-3\cdot(5\ s)^2+4=179\ m.$$

Then, we can find the displacement of the object between $t=0\ s$ and $t=5\ s$:

$$d=d_2-d_1=179\ m-4\ m=175\ m.$$

Unfortunately, from the definition of the question we don't know the force that applied to the object. Let's suppose that the force of 15 N applied to the object. Then, we can find the work done on the object by the force:

$$W=Fd=15\ N\cdot175\ m=2625\ J.$$

5) Unfortunately, we don't know the distance that the block traveled to the right under the action of the horizontal force. Let's suppose that the distance traveled by the block equals 10 meters.

Then, we can find the speed of the block from the work-kinetic energy theorem:

$$KE_f-KE_i=W_{net},$$ $$\dfrac{1}{2}mv_f^2=(F-\mu_kmg)d,$$ $$v_f=\sqrt{\dfrac{2(F-\mu_kmg)d}{m}},$$ $$v_f=\sqrt{\dfrac{2\cdot(20\ N-0.33\cdot5\ kg\cdot9.8\ \dfrac{m}{s^2})\cdot10\ m}{5\ kg}}=3.91\ \dfrac{m}{s}.$$

We can find the acceleration of the block from the Newton's Second Law of Motion:

$$F-\mu_kmg=ma,$$ $$a=\dfrac{F-\mu_kmg}{m},$$ $$a=\dfrac{20\ N-0.33\cdot5\ kg\cdot9.8\ \dfrac{m}{s^2}}{5\ kg}=0.77\ \dfrac{m}{s^2}.$$