- #1
FaNgS
- 91
- 0
(1)An elevator is being lifted up and elevator shaft at a constant speed by a steel cable. All frictional effects are negligible.
In this situation, forces on the elevator are such that?
1. the upward force by the cable is greater than the downward force of gravity.
2. the upward force by the cable is smaller than the downward force of gravity.
3. the upward force by the cable is equal to the downward force of gravity.
4. None of these. (The elevator goes up because the cable is being shortened, not because an upward force is exerted on the eleva-
tor by the cable.)
5. the upward force by the cable is greater than the sum of the downward force of gravity and a downward force due to the air.
Working: my choice is 1, my thought is that since its moving upwards it must mean that the upward force is greater than the downward forces...i neglected 5 since it talks of downward force due to air which is some kind of friction force and the question says to ignore...my choice 1 turned out wrong any idea what's the correct one??
(2)A toy car is given a quick push so that it rolls up an inclined ramp. After it is released, it rolls up, reaches its highest point and rolls back down again. Friction is so small it can be ignored. What net force acts on the car?
1. Net constant force up the ramp
2. Net decreasing force up the ramp
3. Net increasing force down the ramp
4. Net constant force down the ramp
5. Net force of zero
6. Net increasing force up the ramp
7. Net decreasing force down the ramp
Working: I'm confused between them so i did a process of elimination and canceled 5 thinking that since its moving downwards there must be a force acting on the toy car i also canceled choices 1, 6 and 7...1 and 6 because it says force acting upwards which i think isn't possible since the car is moving down the ramp and 7 because if its moving down the ramp the force should be increasing right?
(3)(a)A mass of 1.4 kg lies on a frictionless table, pulled by another mass of 3.1 kg under the influence of Earth's gravity. The acceleration of gravity is 9.8 m/s^2. What is the magnitude of the acceleration a of the two masses? Answer in units of m/s^2.
Working:
For the 1.4kg mass: N(normal force) -1.4g=0
N=1.4g
T(tension)=1.4g
For the 3.1kg mass:
T(tension)-3.1g=-3.1a using T=1.4g and subsituting in
1.4g-3.1g=-3.1a
therefore a=5.374193548 m/s^2
i'm not sure at all if this method is right or if there's something I'm missing
(b) T, a and g represent positive quantities. Which equation is correct?
1. T + (1.4 kg) g = (1.4 kg) a
2. T - (1.4 kg) g = (3.1 kg) a
3. (3.1 kg) g ¡ T = (3.1 kg) a
4. T - (3.1 kg) g = (3.1 kg) a
5. T - (3.1 kg) g = (1.4 kg) a
6. (3.1 kg) g - T = (1.4 kg + 3.1 kg) a
7. (3.1 kg) g - T = (1.4 kg) a
8. T - (3.1 kg) g = (1.4 kg + 3.1 kg) a
9. T - (1.4 kg) g = (1.4 kg) a
10. T + (1.4 kg) g = (3.1 kg) a
Working: i chose eqn4 since that's exactly what i got for part (a) but since there's a possiblity of (a) being wrong I'm not sure
In this situation, forces on the elevator are such that?
1. the upward force by the cable is greater than the downward force of gravity.
2. the upward force by the cable is smaller than the downward force of gravity.
3. the upward force by the cable is equal to the downward force of gravity.
4. None of these. (The elevator goes up because the cable is being shortened, not because an upward force is exerted on the eleva-
tor by the cable.)
5. the upward force by the cable is greater than the sum of the downward force of gravity and a downward force due to the air.
Working: my choice is 1, my thought is that since its moving upwards it must mean that the upward force is greater than the downward forces...i neglected 5 since it talks of downward force due to air which is some kind of friction force and the question says to ignore...my choice 1 turned out wrong any idea what's the correct one??
(2)A toy car is given a quick push so that it rolls up an inclined ramp. After it is released, it rolls up, reaches its highest point and rolls back down again. Friction is so small it can be ignored. What net force acts on the car?
1. Net constant force up the ramp
2. Net decreasing force up the ramp
3. Net increasing force down the ramp
4. Net constant force down the ramp
5. Net force of zero
6. Net increasing force up the ramp
7. Net decreasing force down the ramp
Working: I'm confused between them so i did a process of elimination and canceled 5 thinking that since its moving downwards there must be a force acting on the toy car i also canceled choices 1, 6 and 7...1 and 6 because it says force acting upwards which i think isn't possible since the car is moving down the ramp and 7 because if its moving down the ramp the force should be increasing right?
(3)(a)A mass of 1.4 kg lies on a frictionless table, pulled by another mass of 3.1 kg under the influence of Earth's gravity. The acceleration of gravity is 9.8 m/s^2. What is the magnitude of the acceleration a of the two masses? Answer in units of m/s^2.
Working:
For the 1.4kg mass: N(normal force) -1.4g=0
N=1.4g
T(tension)=1.4g
For the 3.1kg mass:
T(tension)-3.1g=-3.1a using T=1.4g and subsituting in
1.4g-3.1g=-3.1a
therefore a=5.374193548 m/s^2
i'm not sure at all if this method is right or if there's something I'm missing
(b) T, a and g represent positive quantities. Which equation is correct?
1. T + (1.4 kg) g = (1.4 kg) a
2. T - (1.4 kg) g = (3.1 kg) a
3. (3.1 kg) g ¡ T = (3.1 kg) a
4. T - (3.1 kg) g = (3.1 kg) a
5. T - (3.1 kg) g = (1.4 kg) a
6. (3.1 kg) g - T = (1.4 kg + 3.1 kg) a
7. (3.1 kg) g - T = (1.4 kg) a
8. T - (3.1 kg) g = (1.4 kg + 3.1 kg) a
9. T - (1.4 kg) g = (1.4 kg) a
10. T + (1.4 kg) g = (3.1 kg) a
Working: i chose eqn4 since that's exactly what i got for part (a) but since there's a possiblity of (a) being wrong I'm not sure