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