1. How far (in meters) must a 600 kg pile driver fall, if it is to do 14000 J of work?
2. A spring whose spring constant is 150 N/m is compressed 0.800 m. What speed (in meters/second) can it give to a 0.600 kg ball when released?
3. A 550 N crate rests on the floor. How much work in joules is required to move it at constant speed 4.0 m along the floor, against a friction force of 140 N?
4. What is the minimum work in joules needed to push a 1710 kg car 74.0 m up a 12.5 degree incline? (Ignore friction.)
5. How much work in joules must be done to stop a 3000 kg car traveling at 150 km/hr?
6. A 1620 kg car rolling on a horizontal surface has a speed of 70 km/hr when it strikes a horizontal coiled spring and is brought to rest in a distance of 5.5 m. What is the spring constant (in N/m) of the spring? (Ignore nonconservative forces such as friction.)
7. A spring has a spring constant "k" of 160 N/m. How much must this spring be compressed (in meters) to store 70 J of energy?
8. Tarzan is running at a top speed of 5.0 m/s and grabs a vine hanging vertically from a tall tree in the jungle. How high (in meters) can he swing upward?
9. An object slides down a frictionless 33 degree incline whose vertical height is 87.0 cm. How fast is it going in meters/second when it reaches the bottom?
10. A 33 kg child descends a slide 7.8 m high and reaches the bottom with a speed of 3.30 m/s. How much thermal energy (in joules) due to friction was generated in this process?
11. Two railroad cars, each of mass 6800 kg and traveling 80.0 km/hr, collide head-on and come to rest. How much thermal energy in joules is produced in this collision?
See answer below. I am not going to do all these problems for you. I already know how.
To solve these physics problems, we can make use of various formulas and principles. Here's a step-by-step explanation of how to solve each problem:
1. Work is equal to the product of force applied and the distance moved in the direction of the force. In this case, the work is given as 14000 J, and we are looking to find the distance. The formula for work is given by W = F * d. Since the pile driver falls vertically, the force is equal to the weight of the pile driver, which is given as mass * acceleration due to gravity. So, F = m * g. Rearranging the formula, we get d = W / (m * g). Plugging in the given values, we get d = 14000 J / (600 kg * 9.8 m/s²).
2. The potential energy stored in the compressed spring is equal to the work done on the ball. This energy is then transformed into kinetic energy when the ball is released. The formula for potential energy is given by PE = (1/2) * k * x², where k is the spring constant and x is the compression distance. Equating the potential energy to the kinetic energy (KE = (1/2) * m * v²), we can solve for v. Plugging in the given values, we get v = √((2 * PE) / m).
3. The work done on the crate against the friction force is equal to the force multiplied by the distance moved. The formula for work is given by W = F * d. Plugging in the given values, we get W = 140 N * 4.0 m.
4. The work done to push the car up the incline is equal to the change in gravitational potential energy. The formula for work is given by W = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height of the incline. Plugging in the given values, we get W = 1710 kg * 9.8 m/s² * (74.0 m * sin(12.5°)).
5. The work done to stop the car is equal to the change in kinetic energy. The formula for work is given by W = (1/2) * m * v², where m is the mass and v is the velocity of the car. Plugging in the given values, we get W = (1/2) * 3000 kg * (150000 m/3600 s)².
6. The energy stored in the spring is equal to the work done on the car to bring it to rest. The formula for potential energy stored in a spring is given by PE = (1/2) * k * x², where k is the spring constant and x is the distance compressed. Equating the potential energy to the initial kinetic energy of the car, we can solve for k. Plugging in the given values, we get k = (1620 kg * (70 km/hr)^2) / (2 * 5.5 m²).
7. The potential energy stored in the spring is given by PE = (1/2) * k * x². Rearranging the formula, we get x = √((2 * PE) / k). Plugging in the given values, we get x = √((2 * 70 J) / 160 N/m).
8. The maximum height Tarzan can swing upward is determined by the conservation of mechanical energy. At the highest point, all of Tarzan's initial kinetic energy is converted into gravitational potential energy. The formula for gravitational potential energy is given by PE = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the maximum height. Plugging in the given values, we get h = (5.0 m/s)² / (2 * 9.8 m/s²).
9. Assuming no energy losses due to friction, the total mechanical energy of the object (potential + kinetic) remains constant. Therefore, the potential energy at the top of the incline is equal to the kinetic energy at the bottom. The formula for potential energy is given by PE = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the vertical height. Equating the potential energy to the kinetic energy (KE = (1/2) * m * v²), we can solve for v. Plugging in the given values, we get v = √(2 * g * h * sin(33°)).
10. The thermal energy due to friction is equal to the mechanical energy lost by the child's descent. The formula for mechanical energy is given by ME = PE + KE, where ME is the mechanical energy, PE is the potential energy, and KE is the kinetic energy. The thermal energy can be obtained by subtracting the mechanical energy at the bottom (ME_bottom) from the mechanical energy at the top (ME_top). Plugging in the given values, we get thermal energy = ME_top - ME_bottom.
11. In an inelastic collision, the kinetic energy before the collision is equal to the thermal energy generated during the collision. The formula for kinetic energy is given by KE = (1/2) * m * v², where m is the mass and v is the velocity. Plugging in the given values, we get thermal energy = (1/2) * 6800 kg * (80 km/hr)^2 / (1000 m/km)^2.