Question 1 of 7 0.5 Points
A 6 N force acts on a 0.3-kg object for 0.3 seconds. By how muchdoes the object's momentum change (in kg-m/s)?

Question 2 of 7 0.5 Points
A 0.4-kg object is traveling to the right (in the positivedirection) with a speed of 1 m/s. After a 0.2 s collision, theobject is traveling to the left at 3 m/s. What is the magnitude ofthe impulse (in N-s) acting on the object? The answer must bepositive.

Question 3 of 7 1.0 Points
The force shown in the attached figure is the net eastward forceacting on a ball. The force starts rising at t = 0.020 s, fallsback to zero at t = 0.062 s, and reaches a maximum force of 55 N atthe peak. Determine with an error no bigger than 25% (high or low)the magnitude of the impulse (in N-s) delivered to the ball. Hint:Do not use J = F?t. Look at the figure. Find the area of a nearlyequally sized triangle.

Question 4 of 7 1.0 Points
A bat hits a moving baseball. If the bat delivers a net eastwardimpulse of 0.6 N-s and the ball starts with an initial horizontalvelocity of 3.8 m/s to the west and leaves with a 5.2m/s velocityto the east, what is the mass of the ball (in grams)?
Part 5 of 7 -

Question 5 of 7 0.5 Points
The force sensor measures the force on the sensor due to thebumper, but the cart's momentum change arises from the force on thecart due to the bumper. Which of the following facts are needed toassert that the magnitude of these two forces are nearly equal atall times.
1. Their magnitudes differ by the magnitude of the net force on thebumper.
2. The force of friction is small.
3. The net force on the bumper is small.
4. The cart's weight is canceled by the the normal force exerted bythe track.

A. 1, 3

B. 2, 4

C. 3, 4

D. 1, 2

E. 1, 4

F. 1, 2, 3, 4

G. 2, 3

Part 6 of 7 -

Question 6 of 7 1.0 Points
A cart traveling at 0.3 m/s collides with stationary object. Afterthe collision, the cart rebounds in the opposite direction. Thesame cart again traveling at 0.3 m/s collides with a differentstationary object. This time the cart is at rest after thecollision. In which collision is the impulse on the cartgreater?

A. The second collision.

B. Cannot be determined without knowing the mass of the cart.

C. The impulses are the same.

D. The first collision.

E. Cannot be determined without knowing the rebound speed of thefirst collision. Reset Selection

Part 7 of 7 -

Question 7 of 7 0.5 Points
The force vs. time graphs for two collisions are shown. Whichcollision has the largest impulse?

A. The impulse for collision 2 since it lasts longer.

B. Cannot be determined since the masses are unknown.

C. The impulse for collision 1 since it has the larger maximumforce.

D. Cannot be determined since the initial an final speeds areunknown.

E. The impulses are the same.

A boy in a wheelchair (total mass 50.5 kg) has a speed of 1.40 m/s at the crest of a slope 2.40 m high and 12.4 m long. At the bottom of the slope, his speed is 6.50 m/s. Assume air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N. Find the work he did in pushing forward on his wheels during the downhill ride.

2 Answers

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electron1
Lv 7

4 years ago

There are two forces that cause the wheelchair to accelerate. These two forces are the component of the wheelchair’s weight that is parallel to the hill and the force the boy exerts. The net force on the wheelchair is equal to the sum of these two forces minus the friction force. Let’s use the following equation to determine the acceleration.

vf2 = vi^2 + 2 * a * d

6.5^2 = 1.4^2 + 2 * a * 12.4

24.8 * a = 40.29

a = 40.29 ÷ 24.8

This is approximately 1.62 m/s^2.

Net force = 50.5 * 40.29 ÷ 24.8 = 2034.645 ÷ 24.8

This is approximately 82 N.

Force parallel = 50.5 * 9.8 * 2.40 ÷ 12.4 = 1187.76 ÷ 12.4

This is approximately 95.8 N. This force and the force that the boy exerts are the forces that cause the wheelchair to accelerate.

Net force = F + (1187.76 ÷ 12.4) – 41

Set the two net forces equal to each other.

F + (1187.76 ÷ 12.4) – 41 = 2034.645 ÷ 24.8

F = 41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)

F is approximately 27.3 N

Work = [41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)] * 12.4 = 337.9625 J
JupiterSky
Lv 6

4 years ago

His change in kinetic energy is due to gravitational- and his own work,

while the friction does negative work.

∆K = ∆U +∆W + ∆Wf

½m(Vf²−Vi²) = mgh + ∆W + Ff*d

∆W = ½m(Vf²−Vi²) − mgh − Ff*d

∆W = ½*50.5(6.5²−1.4²) − (50.5*9.8*2.4) − (−41*12.4)

∆W = 508 J

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asked in Science & MathematicsPhysics · 8 years ago