Homework 6 Problems: Energy and Work 1. [10 points] A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25° below the horizontal. The force is just sufficient to overcome various frictional forces so the cart moves at constant speed. a. Find the work done by the shopper as she moves down a 50.0-m length aisle. b. What is the net work done on the cart? Explain why. c. The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesn’t change, would the shopper’s applied force be larger, smaller or the same? What about the work done on the cart by the shopper? Answer: (a) As the shopper moves down the 50.0-m length aisle, the work done by the shopper is | || | ( )( ) ( ) (b) Because the cart is moving with a constant speed with a constant direction, the cart is not accelerating. This means that there is no net force operating on the cart and thus the net work must be zero. (c) The applied force would be the same since the cart does not accelerate. However, because the force is directed parallel to the displacement, the work done on the cart by the shopper must increase. 2. [10 points] A daredevil on a motorcycle leaves the end of a ramp with a speed of 35.0 m/s as in the figure below. If this speed is 33.0 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Answer: Assuming that air resistance is negligible on the object, then mechanical energy is conserved for this system. Designating the end of the ramp as , the maximum height can be determined using conservation of mechanical energy Solving for the maximum height gives ( ) ( ( ) ) 3. [15 points] A 7.80-g bullet moving at 575 m/s penetrates a tree trunk to a depth of 5.50 cm. a. Use work and energy considerations to find the average frictional force that stops the bullet. b. Assuming the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving. Answer: (a) To determine the average frictional force, we use the work-energy theorem ( ) ( ) Solving for the frictional force gives ( ) ( ) [( ( ) ( ) ] ) (b) If the frictional force is constant, then the net impulse acting on the bullet will cause a change in linear momentum. This can be written as Solving for time elapsed gives ( )( ) 4. [15 points] A 50.0-kg projectile is fired at an angle of 30.0⁰ above the horizontal with an initial speed of 120.0 m/s from the top of a cliff 142 m above level ground. a. What is the initial total mechanical energy of the projectile? b. Suppose the projectile is traveling 85.0 m/s at its maximum height of 427 m. How much work has been done on the projectile by air resistance? c. What is the speed of the projectile immediately before it hits the ground if air resistance does 1.5 times as much work on the projectile when it’s going down as it did when it was going up? Answer: (a) The initial total mechanical energy of the projectile is the sum of its kinetic energy and its gravitational potential energy. Therefore, we have ( )( ( ) )( )( ) (b) In the presence of air resistance, the mechanical energy of the projectile will decrease as mechanical energy leaves the system. Therefore, the change in mechanical energy is due to the work done by the frictional force of air resistance. The work done on the projectile by air resistance is given by ( ( (c) Let’s define ( ) ) [( ) )( ( ) ( ) ] )[ ] as the work done by air resistance when the projectile goes up and as the work done by air resistance when the projectile goes down. If the work done by air resistance does 1.5 times as much work on the projectile when it’s going down as it did when it was going up, then ( ) As the projectile goes down towards the ground, air resistance will do work on the projectile, causing mechanical energy to leave the system. Therefore, we can use the work-energy theorem to find the speed of the projectile. Here, we set yf = 0 m at the ground, yi = 427 m as the maximum height, and at maximum height v0 = 85.0 m/s. Using the work-energy theorem to obtain the final velocity gives: ( ) ( √ √ √ ( ) ( [( ( ) ) ( √ ) ) ) ( )( ) ( )( )( )] Bonus: [7 points] Two blocks, A and B (with mass 50 kg and 100 kg, respectively), are connected by a string. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between block A and the incline is µk = 0.25. Determine the change in the kinetic energy of block A as it moves from C to D, a distance of 20 m up the incline (and block B drops downward a distance of 20 m) if the system starts from rest. Answer: The work-energy theorem states that the change in kinetic energy of block A is equal to the net work done on block A. Because both blocks are connected by a string, then both blocks will have the same magnitude of acceleration and thus will have the same change in kinetic energy. The gravitational force, the kinetic friction force, and the tension force do work on the system. This can be written as We note that the tension force is directly related to the gravitational force on block B, resulting in the acceleration of the entire system. Therefore, we can rewrite the work-energy theorem as Since the gravitational potential energy associated with an object is the negative of the work done by the force of gravity in moving the object, we can rewrite the work-energy theorem in terms of the gravitational potential energy Using the definition of gravitational potential energy and kinetic friction gives us Here, refers to the vertical distance traveled by block A and refers to the vertical distance traveled by block B. Using the schematic diagram above, we see that and . To determine the normal force n, we can use a free body diagram, given below. θ mAg From the free body diagram above, we see that the system is given by ( ( )( )( )( . Therefore, the net work done on ( ) )( ) )( ) ( )( )( ) Since both blocks are connected through by a string, they will have the same acceleration and thus, the same final velocity given by √ ( √ √ ) Therefore, the change of kinetic energy of block A is given by ( )( )

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