Question

A 12.0 kg block rests on an inclined plane. The plane makes an angle of 31.0° with the horizontal, and the coefficient of friction between the block and the plane is 0.158. The 12.0 kg block is tied to a second block (mass = 38.0 kg) which hangs over the end of the inclined plane after the rope passes over an ideal pulley. (a) What is the acceleration of each of the two blocks, and (b) what is the tension in the rope?

Answer 1

The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.

Step-by-step explanation:

Given that,

Mass = 12.0 kg

Angle = 31.0°

Friction coefficient = 0.158

Mass of second block = 38.0 kg

Using formula of frictional force

`f_(\mu) = \mu N`....(I)

Where, N = normal force

`N = mg\cos\theta`

Put the value of N into the formula

`N =12*9.8*\cos 31^(\circ)`

`N=100.80\ N`

Put the value of N in equation (I)

`f_(mu)=0.158*100.80`

`f_(mu)=15.9264\ N`

Now, Weight of second block

`W = mg`

`W=38.0*9.8`

`W=372.4\ N`

The horizontal force is

`F = mg\sintheta`

`F=12*9.8*\sin 31^(\circ)`

`F=60.5684\ N`....(II)

(I). We need to calculate the acceleration

`a=m_(2)g-(f_(\mu)+mg\sin\theta)/(m_(1)+m_(2))`

`a=(372.4-(15.9264+60.5684))/(12+38)`

`a=5.92\ m/s^2`

(II). We need to calculate the tension in the rope

`m_(2)g-T=m_(2)a`

`-T=38*5.92-38*9.8`

`T=147.44\ N`

Hence, The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.