Answer to Question #192321 in Physics for Nafiyad

Question

1.A 2kg block is moved by a slope inclined at angle of 37° by a horizontal force of 45N.the block moves a distance of 2m up the plane.if µ=0.2 find

A.the work done by the applied force

B.the work done by gravity

C.the work done by friction

D.the change in kinetic energy of the block

2.A ball of mass 3 kg starts from rest at point p and travels along the track 4m.find the velocity at points Q and R if

A.the track is frictionless

B.the friction force is 20N

Answer 1

1.A.

$W_1=Fs\cos37°=45\cdot(\frac{2}{\sin37°})\cdot\cos37°=119.43\ (J)$ . Answer

1.B.

$W_2=Fs\cos180°=mg\cdot 2\cdot\cos180°=2\cdot9.8\cdot2\ \cdot \cos180°=-39.2\ (J)$ . Answer

1.C.

$W_3=F_fs\cos180°=\mu Ns\cos180°$

$N=mg\cos37+F\sin37°=2\cdot 9.8\cdot \cos37+45\cdot\sin37°=28.65\ (N)$

$W_3=\mu Ns\cos180°=0.2\cdot28.65\cdot\frac{2}{\sin37°}\cdot\cos180°=-19.04\ (J)$ . Answer

1.D.

$\Delta KE=W_1-W_2-W_3=119.43-39.2-19.04=61.2\ (J)$ . Answer

2.A.

Assume that the angle of inclination of the track is $\alpha=50°$

$F=ma\to a=F/m=mg\sin50°/m=g\sin50°=9.8\cdot\sin50°=7.5\ (m/s^2)$

$v=\sqrt{2sa}=\sqrt{2\cdot 4\cdot 7.5}=7.75\ (m/s)$ . Answer

2.B.

$F=ma\to a=(F-F_f)/m=(mg\sin50°-20)/3=0.84\ (m/s^2)$

$v=\sqrt{2sa}=\sqrt{2\cdot 4\cdot 0.84}=2.6\ (m/s)$ . Answer