1.A small truck of mass 8000kg is being passed on the freeway by a car of mass 1600kg. If at the point of passing the car is traveling at 100kmh and has the same kinetic energy as the truck, what is the truck's velocity?

2.A toy car of mass 100g is powered by a spring mechanism. If the energy transfer is 100% efficient and the potential energy of the spring at the moment the car is released is 2 J, calculate the speed with which the car can travel.

3.Jane threw a dart of mass 50g into a block of soft wood. The block of mass 1.0kg was suspended freely by a length of fishing line before being hit by the dart. Andrew observed that the block with the dart in it swung to one side and rose a vertical distance of 40cm. Calculate the speed of the dart when it hit the block.

4.Alan had difficulty pushing a peg of mass 200g which was used to hold a ripe to his tent into the ground so he decided to jump onto the peg. He jumped to a height of 30cm and landed vertically on top of the peg which was 15cm above the ground. To his dismay the peg only moved 9 cm into the ground.

If Alan has a mass of 45 kg, find
a.his energy just before impact with the peg,

b.his velocity just before impact with the peg,

c.the velocity of the peg after impact if Alan/s velocity after the impact os the same as that of the peg,

d.the energy loss due to the impact,

e.the average force of friction due to the ground which acted on the peg.

38.During a physics experiment to measure the power necessary to run up a flight of stairs, Maire recorded a time of 10s for Shirley to run up the stairs. The two girls also recorded the following data:

Shirley's mass=50kg
number of steps climbed=30
height of each step=25cm
Calculate
a.the work done by Shirley
b. her power output on reaching the top of the stairs

39. Sam drives his car of mass 1500 kg up a 10 degree hill. Given that the total resistance due to friction is 200N and that he is driving the car at a constant speed of 15ms-1, calculate the power developed by his car's engine.

1. A small truck of mass 8000kg is being passed on the freeway by a car of mass 1600kg. If at the point of passing the car is traveling at 100 kmh and has the same kinetic energy as the truck, what is the truck's velocity?


I'm sure you wouldn't want me to do all of your problems, and deprive you of your learning experience, so I'll only do this one.

Kinetic energy, e, is given by:

e\,=\,\frac 12 mv^2

where v is the velocity, m is the mass, and e is the energy.

e_{truck}\,=\,\frac 12 8000\,kg\,v^2

e_{car}\,=\,\frac 12 1600\,kg\,(100\,kmh)^2

we want the point where e_{car}=e_{truck} ("has the same kinetic energy as the truck") so

\frac 12 8000\,kg\,v_{truck}^2 \,= \,\frac 12 1600\,kg\,(100\,kmh)^2

80v_{truck}^2=16 \,\times\, 10000\,kmh^2

v_{truck}^2=2000\,kmh^2

v_{truck}=\sqrt{2000}\,kmh \approx 44.72\,kmh

For each problem, simply find the formula that represents the energy of the system or the force from the system. I'm sure you'll have no problem finding the equation for the spring and potential energy is given by e=mgh m=mass, h=height, and g is the acceleration due to gravity (9.80665 meters per second per second)

Can you help me with question 2,4 and 39? I managed to solve the others. Thanks :D

2,4, and 39, and I already did #1. So, you did #3 and #38.

#2. Hooke's law says

F_s=kx

where k is the spring constant and x is the displacement of the spring. In this case, we're told Fs = 2 Joules. This is the potential energy. We assume, therefore (because the problem says 100% efficient) that it all goes into kinetic energy

E_s = E_k=2\, joules = \frac 12 mv^2

E_s = E_k=2\, joules = \frac 12 0.1\, kg\, v^2

1 Joule is equivalent to 1 kg-m^2/sec^2. So, you solve for v. and you will get a value in meters/sec.
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4.
Peg's mass = 200 g.
Alan's mass = 45 kg
Alan's height = 30 cm
Top of Peg height = 15 cm

E_a = mgh

where Ea is the potential energy of Alan. M is the mass (of Alan). H is the height that Alan came down. (30 cm - 15 cm= 15 cm). Therefore, the potential energy when Alan was 30 cm in the air was

E_a=45\,kg\,\times\, 9.80665\,\frac {m}{sec^2}\,\times\,0.15\,m

This is the answer to "a" since all potential energy is converted to kinetic energy.

Looking at the units, Ea will be in kg m^2/sec^2 which, as I mentioned above, is a Joule. That energy drove the peg 6 cm into the ground.

b) The answer to "b" is found by the same method I used in problem #1. Set the potential energy equal to the kinetic energy (1/2 mv^2) and solve for v.

c) The answer to "c" is found by the same method as b, except that the mass has changed. So

\frac 12 m_av_a^2 = \frac 12 m_p v_p^2

You know alan's mass, the peg's mass, and alan's velocity, so you solve for the peg's velocity.

d) The energy loss is the entire kinetic energy.

e) The peg decelerates from whatever you calculated in c) to a velocity of zero. All of the energy goes into the ground. The stopping distance was 9 cm Energy is force times distance.

W=\frac 12 m \Delta v^2 (Δv = v2-v1, but v2=0 so the velocity is the velocity calculated in c).

W=fd

You know the distance (9 cm = 0.09 meters), you know the energy, so solve for the force. D is in meters; e is in joules, so f will be in newtons (kg m/sec^2).
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39. Sam drives his car of mass 1500 kg up a 10 degree hill. Given that the total resistance due to friction is 200N and that he is driving the car at a constant speed of 15ms-1, calculate the power developed by his car's engine.

The resistance due to friction, 200 Newtons, is a force. It's going to be added to the other forces.

Power = Work / Time
Work = Force x Distance

You have a vector pointed 10 degrees upward from the horizontal. The vertical component of that vector is the length of the vector times the sine of the angle (10 degrees).

F_v=1500 \, kg \, \times 9.80665\, \frac {m}{sec^2}\,sin(10) = 2554.4\,N

You need to add that to the 200 Newtons of resistance force. You know the speed of the car so you can calculate the total power required.

Thanks man you are a legend. THX so much