1. A weight lifter lifts a set of weights a vertical distance of 3.85 m. If a constant net force of 350 N is exerted on the weights, what is the net work done on the weights? (Show all work and include units of measure)
2. A car has a kinetic energy of 4.32 x 10^5 J when traveling at a speed of 23 m/s. What is its mass? (Show all work and include units of measure)
3. A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 45 N. How far must the student be pushed, starting from rest, so that her final kinetic energy is 352 J? (Show all work and include units of measure)
Help me please
1. work = force * distance
2. K.E. = 1/2 * m * v^2 ... m = 2 * K.E. / v^2
3. same as 1 ... work and energy are equivalent ... distance = K.E. / force
Sure! I'd be happy to help you with these physics problems. Let's go through each question step by step:
1. To find the net work done on the weights, you can use the formula: Work = Force × Distance. In this case, the force is a constant net force of 350 N, and the distance is 3.85 m. So, the net work done on the weights is:
Work = 350 N × 3.85 m = 1347.5 Joules.
Therefore, the net work done on the weights is 1347.5 Joules.
2. To find the mass of the car, you need to use the formula for kinetic energy: Kinetic Energy = (1/2) × Mass × Velocity^2. In this case, the kinetic energy is 4.32 x 10^5 J and the velocity is 23 m/s. Rearranging the formula, we get:
Mass = (2 × Kinetic Energy) / (Velocity^2)
= (2 × 4.32 x 10^5 J) / (23 m/s)^2
= (8.64 x 10^5 J) / (529 m^2/s^2)
= 1633.63777324 kg (approximately)
Therefore, the mass of the car is approximately 1633.64 kg.
3. To find the distance the student must be pushed, you can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. The work done by the pushing force is equal to the change in kinetic energy, so:
Work = Change in Kinetic Energy = Final Kinetic Energy - Initial Kinetic Energy
In this case, the final kinetic energy is 352 J, and the initial kinetic energy is zero (since the student starts from rest). So, the work done is:
Work = 352 J - 0 J = 352 J
Now, we can use the formula for work to find the distance. The work done is equal to the force multiplied by the distance, so:
Work = Force × Distance
352 J = 45 N × Distance
Divide both sides of the equation by 45 N to solve for the distance:
Distance = 352 J / 45 N = 7.8222 m (approximately)
Therefore, the student must be pushed a distance of approximately 7.82 meters.
I hope that helps! Let me know if you have any further questions.