Work
Work
The work done on an object by a constant force is
W = Fd = Fcos d where F is the magnitude of the force, d is the object’s displacement, and is the angle between the direction of the force and the displacement . Solving simple problems requires substituting values into this equation. More complex problems, such as those involving friction, often require using Newton’s second law to determine forces.
Example:
An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force of 120 N on the sled by pulling on the rope.
(a) How much work does he do on the sled if the rope is horizontal to the ground and he pulls the sled 5.00 m?
(b) How much work does he do on the sled if = 30.0° and he pulls the sled the same distance?
Suppose that the coefficient of kinetic friction between the loaded 50.0-kg sled and snow is 0.200.
(c) The Eskimo again pulls the sled 5.00 m, exerting a force of 120 N at an angle of 0°. Find the work done on the sled by friction, and the network.
(d) Repeat the calculation if the applied force is exerted at an angle of 30.0° with the horizontal.
Kinetic Energy and the Work–Energy Theorem
The kinetic energy KE of an object of mass m moving with a speed v is defined by
KE = ½ mv2
SI unit: joule ( J ) = kgm2/s2
The work–energy theorem states that the net work done on an object of mass m is equal to the change in its kinetic energy, or
Wnet = KEf - KEi = Fd where the change in the kinetic energy is due entirely to the object’s change in speed.
Example:
The driver of a 1000 kg car traveling on the interstate at 35.0 m/s slams on his brakes to avoid hitting a second vehicle in front of him, which had come to rest because of congestion ahead. After the brakes are applied, a constant friction force of 8000 N acts on the car. Ignore air resistance.
(a) At what minimum distance should the brakes be applied to avoid a
The work done on an object by a constant force is
W = Fd = Fcos d where F is the magnitude of the force, d is the object’s displacement, and is the angle between the direction of the force and the displacement . Solving simple problems requires substituting values into this equation. More complex problems, such as those involving friction, often require using Newton’s second law to determine forces.
Example:
An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force of 120 N on the sled by pulling on the rope.
(a) How much work does he do on the sled if the rope is horizontal to the ground and he pulls the sled 5.00 m?
(b) How much work does he do on the sled if = 30.0° and he pulls the sled the same distance?
Suppose that the coefficient of kinetic friction between the loaded 50.0-kg sled and snow is 0.200.
(c) The Eskimo again pulls the sled 5.00 m, exerting a force of 120 N at an angle of 0°. Find the work done on the sled by friction, and the network.
(d) Repeat the calculation if the applied force is exerted at an angle of 30.0° with the horizontal.
Kinetic Energy and the Work–Energy Theorem
The kinetic energy KE of an object of mass m moving with a speed v is defined by
KE = ½ mv2
SI unit: joule ( J ) = kgm2/s2
The work–energy theorem states that the net work done on an object of mass m is equal to the change in its kinetic energy, or
Wnet = KEf - KEi = Fd where the change in the kinetic energy is due entirely to the object’s change in speed.
Example:
The driver of a 1000 kg car traveling on the interstate at 35.0 m/s slams on his brakes to avoid hitting a second vehicle in front of him, which had come to rest because of congestion ahead. After the brakes are applied, a constant friction force of 8000 N acts on the car. Ignore air resistance.
(a) At what minimum distance should the brakes be applied to avoid a