How Can Physics Solve These Real-World Motion Problems?

[ispyy]

1. A spring has a force constant of 30000 N/m.
How far must it be stretched for its poten-
tial energy to be 47 J?
Answer in units of m.

2. A bobsled slides down an ice track starting
(at zero initial speed) from the top of a(n)
94.5 m high hill.
The acceleration of gravity is 9.8 m/s2 .
Neglect friction and air resistance and de-
termine the bobsled’s speed at the bottom of
the hill.
Answer in units of m/s.

3. An ore car of mass 40000 kg starts from rest
and rolls downhill on tracks from a mine. At
the end of the tracks, 9.4 m lower vertically,
is a horizontally situated spring with constant
4.5 × 105 N/m.
The acceleration of gravity is 9.8 m/s2 .
Ignore friction.
How much is the spring compressed in stop-
ping the ore car?
Answer in units of m.

4. A roller coaster cart ofmassm = 336 kg starts
stationary at point A, where h1 = 27.7 m and
a while later is at B, where h2 = 13.3 m.
The acceleration of gravity is 9.8 m/s2 .
What is the speed of the cart at B, ignoring
the effect of friction?
Answer in units of m/s.

 

[anathema]

1. To find the distance a spring must be stretched for its potential energy to be 47 J, we can use the formula for potential energy of a spring, U = 1/2kx^2, where U is the potential energy, k is the force constant, and x is the distance stretched. Rearranging the formula, we get x = √(2U/k). Plugging in the values, we get x = √(2(47 J)/(30000 N/m)) = 0.115 m. Therefore, the spring must be stretched 0.115 m for its potential energy to be 47 J.

2. To find the speed of the bobsled at the bottom of the hill, we can use the formula for motion with constant acceleration, v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and s is the displacement. Rearranging the formula, we get v = √(2as). Plugging in the values, we get v = √(2(9.8 m/s^2)(94.5 m)) = 43.6 m/s. Therefore, the bobsled's speed at the bottom of the hill is 43.6 m/s.

3. To find the compression of the spring, we can use the formula for potential energy of a spring, U = 1/2kx^2, where U is the potential energy, k is the force constant, and x is the compression distance. We can also use the conservation of energy principle, where the potential energy at the top of the hill is equal to the potential energy at the bottom, plus the kinetic energy. Rearranging the formula, we get x = √(2U/k). Plugging in the values, we get x = √(2(40000 kg)(9.8 m/s^2)(9.4 m)/(4.5 × 10^5 N/m)) = 0.165 m. Therefore, the spring is compressed 0.165 m in stopping the ore car.

4. To find the speed of the cart at point B, we can use the conservation of energy principle, where the potential energy at point A is equal to the potential energy at point B, plus the kinetic energy. Rearranging

 

[kerimek]

1. To determine the distance the spring must be stretched, we can use the formula for potential energy: PE = 1/2 kx^2, where k is the force constant and x is the distance stretched. Rearranging the formula, we get x = √(2PE/k). Plugging in the given values, we get x = √(2*47/30000) = 0.042 m. Therefore, the spring must be stretched 0.042 m for its potential energy to be 47 J.

2. Using the formula for potential energy (PE = mgh) and kinetic energy (KE = 1/2 mv^2), we can set them equal to each other to solve for the final velocity (v) at the bottom of the hill. This gives us v = √(2gh), where g is the acceleration of gravity and h is the height of the hill. Plugging in the given values, we get v = √(2*9.8*94.5) = 43.8 m/s. Therefore, the bobsled's speed at the bottom of the hill is 43.8 m/s.

3. To determine the compression of the spring, we can use the conservation of energy principle where the initial potential energy (PE) of the ore car is equal to the final potential energy of the spring when it stops. This gives us PE = 1/2 kx^2, where k is the spring constant and x is the compression distance. Rearranging the formula, we get x = √(2PE/k). Plugging in the given values, we get x = √(2*mg*h/k), where m is the mass of the ore car, g is the acceleration of gravity, and h is the height of the track. This gives us x = √(2*40000*9.8*9.4/4.5*10^5) = 0.033 m. Therefore, the spring will be compressed 0.033 m when it stops the ore car.

4. Using the conservation of energy principle again, we can set the initial potential energy (PE) of the cart at point A equal to the final potential energy (PE) at point B, plus the kinetic energy (KE) at point B. This gives us PE1 = PE2 + KE2, where PE1