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ScoutFCM

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I have this take home test I'm doing for Physics. I did all the problems and was just wondering if someone can correct me if I'm wrong on any of the problems because I'm on the borderline B-/C+ and I could really use your guys' help to check the problems. Thanks a lot.

**=not sure/dont know how to do

1) Two ropes are attached to a 40kg object. The first rope applies a force of 25N and the second 40N. If the two ropes are perpendicular to each other, what is the resultant acceleration of the object?

Answer: Fnet= (25N^2 + 40N^2)^1/2 = 47.17N/40kg = 1.18m/s^2

2) An airplane of mass 1.2 x 10^4kg tows a glider of mass .6 x 10^4kg. The airplane propellers provide a net forward thrust of 3.6 x 10^4N. What is the glider's acceleration?

Answer: 3600N = (1200kg+600kg)a= 2m/s^2

3)A baseball batter hits an incoming 40m/s fastball. The ball leaves the bat at 50m/s after a ball-on-bat contact time of 0.030s. What is the force exerted on the .15kg baseball?

Answer: F= (.15kg)(90m/s) / .030s = 450N

4)A 9.0kg hanging weight is connected by a string over a pulley to a 5.0kg block sliding on a flat table. If the coefficient of sliding friction is 0.20, find the tension in the string.

Answer: a= (m2g-(.2)m1g)/(m1+m2) a= (9.0kg)(9.8m/s^2) - (0.20)(5.0kg)(9.8m/s^2) / (9.0kg + 5.0kg) = 5.6 m/s^2

T= (5.0kg)(5.6m/s^2)+9.8N= 37.8

5)A worker pulls a 200-N packing crate at constant velocity across a rough floor by exerting a force F = 55.0N at an angle of 35 degrees above the horizontal. What is the coefficient of kinetic friction of the floor?

Answer: 55cos35N= mK (200N+55sin35N) = 45.05N/231.55N = mK = .195

7) What is the minimum amount of energy required for an 80-kg climber, who is carrying a 20-kg pack to climb Mt Everest, 8,850m high?

Answer: mgyi = (100kg)(9.8m/s^2)(8850m) = 8673kJ

8) A 2.00kg ball has zero kinetic and potential energy. Ernie drops the ball into a 10.0m deep well.

a) Determine the sum of its kinetic and potential energy just before the ball hits the bottom

b) Determine the sum of its kinetic and potential energy after the ball comes to a stop in the mud.

Answer: PE= 2kg x 9.8m/s^2 x -10m = -196J

KE = Vi^2 + 2gh = 2(9.8m/s^2)(10m) = 196J

196J-196J=0

9) A parachutist of mass 50.0kg jumps out of an airplane at a height of 1000m. The parachute deploys, and she lands on the ground with a speed of 5.0m/s. How much energy was lost to air friction during this jump?

Answer:KEf+PEf-KEi-PEi

625J-0+0-49000J

48375J lost.

10) A 1000kg sports car accelerates from zero to 25m/s in 7.5s. What is the average power delivered by the automobile engine?

Answer: KE = 1/2(1000kg)(25m/s)^2 = 312.5kJ

P = KE/t = 312.5kJ/7.5s = 41.7kW

11) A girl and her bike have a total mass of 40kg. At the top of the hill her speed is 5.0m/s. The hill is 10m high and 100m long. If the force of friction as she rides down the hill is 20N, what is her speed at the bottom?

Answer: a= .475m/s^2

10^2+100^2=x^2

x=100.5m

v=Vo^2+2ax

v=(120.47m^2/s^2)^1/2 = 10.98m/s

13) Popeye, of mass 70kg, has just downed a can of spinach. He accelerates quickly and stops Bluto, of mass 700kg, who is charging in at 10m/s. What was Popeye's speed?

Answer: 700kg x 10m/s = 7000N 7000N/70kg=100m/s

15) Object 1 has twice the mass of object 2. Each of the objects has the same magnitude of momentum. Determine the ratio of their kinetic energies.

Answer: object 1:p=(10kg)(5m/s)

object 2:p=(5kg)(10m/s)

v of ob.2 = 2 times v of ob.1

KE=1/2mv^2

ob.1: 1/2(10kg)(5m/s)^2 = 125J

ob.2: 1/2(5kg)(10m/s)^250J

KE2 = 2KE1

**=not sure/dont know how to do

1) Two ropes are attached to a 40kg object. The first rope applies a force of 25N and the second 40N. If the two ropes are perpendicular to each other, what is the resultant acceleration of the object?

Answer: Fnet= (25N^2 + 40N^2)^1/2 = 47.17N/40kg = 1.18m/s^2

2) An airplane of mass 1.2 x 10^4kg tows a glider of mass .6 x 10^4kg. The airplane propellers provide a net forward thrust of 3.6 x 10^4N. What is the glider's acceleration?

Answer: 3600N = (1200kg+600kg)a= 2m/s^2

3)A baseball batter hits an incoming 40m/s fastball. The ball leaves the bat at 50m/s after a ball-on-bat contact time of 0.030s. What is the force exerted on the .15kg baseball?

Answer: F= (.15kg)(90m/s) / .030s = 450N

4)A 9.0kg hanging weight is connected by a string over a pulley to a 5.0kg block sliding on a flat table. If the coefficient of sliding friction is 0.20, find the tension in the string.

Answer: a= (m2g-(.2)m1g)/(m1+m2) a= (9.0kg)(9.8m/s^2) - (0.20)(5.0kg)(9.8m/s^2) / (9.0kg + 5.0kg) = 5.6 m/s^2

T= (5.0kg)(5.6m/s^2)+9.8N= 37.8

5)A worker pulls a 200-N packing crate at constant velocity across a rough floor by exerting a force F = 55.0N at an angle of 35 degrees above the horizontal. What is the coefficient of kinetic friction of the floor?

Answer: 55cos35N= mK (200N+55sin35N) = 45.05N/231.55N = mK = .195

****6) A block is launched up an incline plane. After going up the plane, it slides back down to its starting position. The coefficient of friction between the block and the plane is 0.3. Compare the time for the trip up the plane with the time for the trip down the plane. Determine whether the ratio is greater than, equal to, or less than one.**

Answer: ????????????Answer: ????????????

7) What is the minimum amount of energy required for an 80-kg climber, who is carrying a 20-kg pack to climb Mt Everest, 8,850m high?

Answer: mgyi = (100kg)(9.8m/s^2)(8850m) = 8673kJ

8) A 2.00kg ball has zero kinetic and potential energy. Ernie drops the ball into a 10.0m deep well.

a) Determine the sum of its kinetic and potential energy just before the ball hits the bottom

b) Determine the sum of its kinetic and potential energy after the ball comes to a stop in the mud.

Answer: PE= 2kg x 9.8m/s^2 x -10m = -196J

KE = Vi^2 + 2gh = 2(9.8m/s^2)(10m) = 196J

196J-196J=0

9) A parachutist of mass 50.0kg jumps out of an airplane at a height of 1000m. The parachute deploys, and she lands on the ground with a speed of 5.0m/s. How much energy was lost to air friction during this jump?

Answer:KEf+PEf-KEi-PEi

625J-0+0-49000J

48375J lost.

10) A 1000kg sports car accelerates from zero to 25m/s in 7.5s. What is the average power delivered by the automobile engine?

Answer: KE = 1/2(1000kg)(25m/s)^2 = 312.5kJ

P = KE/t = 312.5kJ/7.5s = 41.7kW

11) A girl and her bike have a total mass of 40kg. At the top of the hill her speed is 5.0m/s. The hill is 10m high and 100m long. If the force of friction as she rides down the hill is 20N, what is her speed at the bottom?

Answer: a= .475m/s^2

10^2+100^2=x^2

x=100.5m

v=Vo^2+2ax

v=(120.47m^2/s^2)^1/2 = 10.98m/s

13) Popeye, of mass 70kg, has just downed a can of spinach. He accelerates quickly and stops Bluto, of mass 700kg, who is charging in at 10m/s. What was Popeye's speed?

Answer: 700kg x 10m/s = 7000N 7000N/70kg=100m/s

15) Object 1 has twice the mass of object 2. Each of the objects has the same magnitude of momentum. Determine the ratio of their kinetic energies.

Answer: object 1:p=(10kg)(5m/s)

object 2:p=(5kg)(10m/s)

v of ob.2 = 2 times v of ob.1

KE=1/2mv^2

ob.1: 1/2(10kg)(5m/s)^2 = 125J

ob.2: 1/2(5kg)(10m/s)^250J

KE2 = 2KE1

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