The Physical World I

Spring 2005
MWF 11:30

Instructor: Dr. Bruce Law, CW 327, Tel: 532-1618.

REVIEW QUESTIONS OF Home Work 4

R3. The force of friction must be equal in magnitude but opposite in direction to the force that you are pushing with. This is because the net force is zero.

R5. You are pushing with a force which is equal in magnitude but opposite in direction to the force of friction so that the net force is zero and you can maintain a constant velocity.

R12. Mass is more fundamental because it is an intrinsic property of an object. Weight varies with location depending upon the acceleration due to gravity eg. for a mass m = 10kg on Earth it`s weight is W = mg = 10 x 10 = 100N. However, in space where g = 0m/s² this objects weight is W = mg = 10 x 0 = 0N whereas its mass still remains 10kg.

R16. W = mg = 1 x 10 = 10N

R23. The acceleration a = F(net)/m therefore if F(net) is tripled then a is also tripled.

R24. Acceleration a = F(net)/m, therefore, for a constant F(net) if m is tripled then the acceleration must decrease by a factor of 3.

R25. a = F(net)/m. If both F(net) and m are tripled then a remains unchanged.

R26. The acceleration and net force point in the same direction.

EXCERISES OF Home Work 4

E1. The net force must be 0N because the velocity is constant.

E2. Yes, for example, when you throw an object up into the air the acceleration due to gravity is constant and equal to 10m/s/s, down. However, the velocity of the object reverses direction on the way up compared with on the way down.

E3. No forces can still be acting, it`s just that the net force is 0N eg. when you stand on the ground your weight (which acts down) is balanced by the support force (which acts up).

E6. No, it is not possible to go around a curve in the absence of a force. When you go around a curve, your direction changes and hence your velocity changes, therefore you are accelerating. According to Newton's 2nd Law of motion a net force must be present to cause this acceleration.

E9. The bear has a weight W = mg = 400 x 10 = 4000N, down. As the bear is sliding at constant velocity it`s net force must be zero i.e. there must be a frictional force of 4000N, up.

E15. W(moon) = mg(moon) = 10 x 10/6 = 16.7N (here I divided by 6 gravity on the moon is only 1/6 that of Earth). W(Earth) = m g(Earth) = 10 x 10 = 100N. The mass m = 10kg on both the moon and Earth.

E33. You have to run the engine in order to balance the frictional force which opposes the motion.

E35. The apple is held at rest, therefore, the net force is 0N. When you drop the apple the net force on the apple is just due to its weight W = 1N. (Note that in the problem they call it a 1-N apple therefore they are referring to its weight (rather than it`s mass which would be in kilograms (kg))).

PROBLEMS OF Home Work 4

P2. Acceleration, a = F/m = 200N/40kg = 5m/s/s in the direction of the force.

P3. Upward force = 300N. Weight of pail W = mg = 20 x 10 = 200N.

Therefore the net force on the pail is F(net) = 300 – 200 = 100 N, up.

The acceleration of the pail is a = F(net)/m = 100N, up/20kg = 5m/s², up.