A 1kg mass is projected down a rough circular track (radius=2m) placed in vertical plane as shown . The speed of the mass at point P is 3m/s and at point Q it is 6m/s. How much work is done on the mass between point P and Q by the force of friction ? (take g=10m/s²)
A. -7.5 J. B. -8.5J. C -6.5J. D. -24J
Answers
According to the figure, there are acting two forces on the mass 1 kg.
One is gravitational force and other is force of friction as circular track is rough.
So,
Total work done on the mass= work done by gravity + work done by force of friction
We know that
work done by gravity, w1= mgh= 1*10*2 ( since h=2, from the given figure)
w1= 20 J
Using work energy theorem, we have
Total work done= change in kinetic energy
w1 + work done by force of friction= 1/2m(v_f)^2-1/2m(v_i)^2
where v_f= final velocity, v_i= initial velocity
Now, plugging the value of v_f= 3m/s², v_i= 6m/s², and w1= 20 Nm
20 + work done by force of friction= 1/2*1*6^2-1/2*1*3^2 =
work done by force of friction= 36/2-9/2 - 20= -13/2= -6.5
Therefore, option C is correct, i.e., -6.5 J.
Answer:
-6. 5 J
Explanation:
The same method as mentioned above.