A large crate filled with physics laboratory equipment must be moved up an incline onto a truck.
1. The crate is at rest on the incline. What can you say about the force of friction acting on the crate?
a. The friction force point up the include
b. The friction force point down the include
c. the firction force is 0
2. A physicist attempts to push the crate up the incline. The physicist senses that if he applies slightly more force the crate will move up the incline but cannot muster enough strength to get the motion started. What can you say now about the force of friction acting on the crate?
a. The friction force point up the include
b. The friction force point down the include
c. the firction force is 0
3. he first physicist gets a second physicist to help. They both push on the crate, parallel to the surface of the incline, and it moves at constant speed up the incline. How does the force exerted by the two physicists on the crate compare with the force of friction on the crate?
5 Answers
-
1. Friction is always opposite to actual or impending motion. On a slope, if a body is resting, the gravity is trying to move it —wards, so the friction must be — the incline.
2. Again, if somebody is trying to push it upwards, and is not able to overcome gravity, then the impending motion is —wards, and friction must be —-wards.
3. It is now actual motion up the incline. So the friction has to be ….wards. Sine the motion upwards must overcome both friction and component of gravity along the incline, and there is no acceleration, the exerted force must be equal to the sum of both.
-
1) the friction force points up the incline because it is resisting the box sliding down.
2)it is pointing down the incline because it is resisting the scientists efforts to push it up.
3) they are the same because it is moving at a constant speed so the net force is zero.
-
The rock is thrown up at 20 m/s at the top the velocity has dropped to 0 When it falls again the velocity will be -20 m/s i.e it has the same speed as before but it will now be down So the CHANGE in velocity is -40 m/s Now as dV = a t -40 = -10 * T T = -40 / -10 = 4 s
-
For #3, the force exerted by the two physicists would be greater than the friction on the crate because it moves at a constant speed up the incline.
-
1=c
2=a
idk bout 3
