PHY-2048C – Spring 2021 SI with Camilo

Exam 2 Practice Test

Disclaimer: This practice test does not necessarily cover all the material that is going to appear on the test, the questions are not the same as the actual test, and should not be used as the sole study guide for the test.

1. A ball is being swung by a rope in a uniform circular motion. The rope is 10 m long and the ball’s speed is a constant 20 m/s. Find the ball’s acceleration and the period.

2. Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force?

Ans. Centripetal force is defined as any net force causing uniform circular motion. The centripetal force is not a new kind of force. The label “centripetal” refers to any force that keeps something turning in a circle. That force could be tension, gravity, friction, electrical attraction, the normal force, or any other force. Any combination of these could be the source of centripetal force, for example, the centripetal force at the top of the path of a tetherball swung through a vertical circle is the result of both tension and gravity.

3. A traffic light weighting 122 N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37∘ And 53∘ with the horizontal. These upper cables are not as strong as the vertical cable, and will break if the tension in either one of them exceed 100N. Will the traffic light remain hanging or will one of the cables break? Draw free body diagram and calculate tensions in all cables.

4. Two masses of 80 kg and 140 kg hang from a rope that runs over a pulley. You can assume that the rope is massless and inextensible, and that the pulley is frictionless. Find the upward acceleration of the smaller mass and the tension in the rope.

5. A box is on a 30-degree inclined ramp. This mass is connected by a string to another mass that is suspended over a massless and frictionless pulley. Mass 1 is 10 kg; Mass 2 is 30 kg. The coefficient of kinetic friction of Mass 1 and the inclined surface is 0.2.

a) Find the acceleration of the system. b) Find the tension in the rope.

cos (30 )=Fg1 yFg1

Fg1 cos (30 )=Fg y1

Fg1sin (30 )=Fgx1

F k=uk∗N

∑❑Fm1x=−Fk+T−Fgx 1=m1a

−uk N+T−Fgx 1=m1a

∑❑Fm1y=N−Fg y 1=0

N=F gy 1=m1gcos (30 )

∑❑Fm2y=F g2−T=m2a

−ukm1gcos (30 )+T−Fg x1=m1a

T=m1a+ukm1gcos (30 )+Fg x1

Fg2−T=m2a

Fg2−m1a−ukm1gcos (30 )−Fgx 1=m2a

Fg2−ukm1gcos (30 )−Fgx 1=(m2+m1 )a

(m2g−ukm1gcos (30 )−m1 gsin (30 ) )(m2+m1 )

=a

(m2g+m1 g (−uk cos (30 )−sin (30 ) ))(m2+m1 )

=a=5.71m / s2

m2g−m2a=T=120 N

6. A waitress shoves a ketchup bottle with mass 0.45 kg to her right along a smooth, level lunch counter. The bottle leaves her hand moving at 2.8m/s, then slows down as it slides because of a constant horizontal friction force exerted on it by the countertop. It slides for 1.0 m before coming to rest. What are the magnitude and direction of the friction force acting on the bottle?

7. A curve has a radius of 50 meters and a banking angle of 15o. What is the ideal, or critical, speed (the speed for which no friction is required between the car's tires and the surface) for a car on this curve?

r = 100 m v = 25 m/s g = 9.8 m/(s^2)

Fnet y=Ncos (θ )−mg=may=m∗0=0

N= mgcos (θ )

Fnetx=Nsin (θ)=Fcentripetal=mac=mV 2

r

Nsin(θ)= mgcos (θ )

sin(θ)=mgtan(θ)=m V2

r

gtan(θ)=V2

r

V=√rgtan (θ)=√(50m)(9.8m /s2) tan(15)=11.458m /s

8. A spring was at its resting position where it is attached to a wall at its left side and a block at its right side as shown below. The spring constant is k= 75N/m. The block is moved leftwards and the spring's right end is displaced from the position of 0.25m to 0.15m where it is held in place. All positions are measured from the origin O of the wall.

What is the spring force exerted on the block?

Assume rightwards as the positive direction.

∆ x=0.15−0.25=−0.1m

F spring=−k ∆x=−(75)∗(−0.1)=7.5N

9. A golf player hits a 0.045kg golfball that is initially at rest, changing its momentum by 4.2 (kg m)/s What is the final speed of the golfball?

pchange=p f−p0

pchange=mV f−0

V f=pchangem

= 4.20.045

=93.333m /s

10. A 0.530-kg basketball hits a wall head-on with a forward speed of 18.0 m/s. It rebounds with a speed of 13.5 m/s. The contact time is 0.100 seconds. (a) determine the impulse with the wall, (b) determine the force of the wall on the ball.

a)

impulse=J=∆ p=m∆v=mv f−mv0=(0.530kg)•(−13.5m /s−18.0m /s)

J=−16.7Ns

b)

J=Ft

F=J / t=(−16.7)/0.1=−167 N

11. A force of 120 N is applied to the front of a sled at an angle of 28.0 above the horizontal so as to pull the sled a distance of 165 meters. How much work was done by the applied force?

Ans. W = F •d cos = 120 N • 165 m cos 28.0 = 17,482.36 J