Answer to Question #82668 in Classical Mechanics for cherry

Question

1. You push your physics book 0.8 m along a horizontal table top with a horizontal push of 4.1 N while the opposing force of friction is 1.9N . How much is the net work done on the book? the answer has the unit of J.

2. A 4 kg watermelon is dropped from rest from the roof of a 6.89 m tall building and feels no appreciable air resistance. Just before it hits the ground, what is the speed of the watermelon (unit is m/s)? g=9.8ms-2

3. Two blocks with different masses m1 and m2 and m1 is larger than m2. They are attached to either end of a light rope that passes over a light, frictionless pulley suspended from the ceiling. The mass are released from rest, and the more massive one starts to descend. After this block has descended 2.6 m, its speed is 2.4 ms-1. If the total mass is 1.4kg, what is the mass of the heavier block m1 (unit is kg)? g=9.8ms-2

Answer 1

Solution

W_net=F_net d

F_net=F-F_fr

W_net=(F-F_fr )d=(4.1-1.9)0.8=1.76 J.

2. A 4 kg watermelon is dropped from rest from the roof of a 6.89 m tall building and feels no appreciable air resistance. Just before it hits the ground, what is the speed of the watermelon (unit is m/s)? g=9.8ms-2

Solution

From the conservation of energy:

(mv^2)/2=mgh

v=√2gh=√(2(9.8)(6.89))=11.6 m/s.

3. Two blocks with different masses m1 and m2 and m1 is larger than m2. They are attached to either end of a light rope that passes over a light, frictionless pulley suspended from the ceiling. The mass are released from rest, and the more massive one starts to descend. After this block has descended 2.6 m, its speed is 2.4 ms-1. If the total mass is 1.4kg, what is the mass of the heavier block m1 (unit is kg)? g=9.8ms-2

Solution

a=(M-m)/(M+m) g.

a=v^2/2s

v^2/2s=(M-m)/(M+m) g=(2M-(M+m))/(M+m) g

〖2.4〗^2/(2(2.6))=(2M-1.4)/1.4 9.8

M=0.78 kg.