# Force and Newton's 2nd law

1. If a cord is pulled upward, making an angle of 27º with respect to the horizon line, produces a force of 365N and can pull a box over the floor with a weight of 55.2 kg at a constant speed of 20.5 centimeters/sec - which is the magnitude of the friction force that opposes the movement of the box?

2. A boat pulls a water-skier at a constant speed of 13.5 meters/seconds. The tension of the cable is 165N - which is the magnitude of the friction force that the water and the air produces against the skier?

3. A parachuter of 72 Kilograms falls at a constant speed of 9.1 meters per second. The parachuter has a mass of 6.5 Kilograms - a) What is the parachuter's weight? b) How big is the ascendant force that the air exercises on the parachuter and on the parachute?

4. A car of 1,270 Kilograms requires a horizontal force of 4770N to accelerate 0.175meters/seconds2 on a horizontal road. Which is the magnitude of the force that opposes said movement?

5. A car of 1,060 kg can accelerate from 0 to 80 kilometers in 9.4 seconds - How much force must should be exercise on the vehicle in order to obtain this acceleration?

6. A car pulls another car that weights 1,730 kilograms, if the car that is pulled must accelerate uniformly from 0 to 2.3 meter per second ... in 10.3 seconds ... How much force must be exercised by the pulling cable?

7. a car of 1570 kg moving at 17.5 meter/second must stop at 94.5 meters. Which is the magnitude of the stopping force? Suppose a steady deceleration.

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Note * denotes multiplication, ^ denotes power

1. If a cord is pulled upward, making an angle of 27º with respect to the horizon line, produces a force of 365N and can pull a box over the floor with a weight of 55.2 kg at a constant speed of 20.5 centimeters/sec - which is the magnitude of the friction force that opposes the movement of the box?

Since the box is moving with constant velocity ( i.e. no net force) , frictional force- horizontal component of force in the cord=0 or frictional force= horizontal component of force in the cord

T= force in the cord= 365 N

horizontal component of force in the cord= T cos 27= 325.2 N =365*0.89101

Answer: magnitude of the friction force that opposes the movement of the box= 325.2 N

2. A boat pulls a water-skier at a constant speed of 13.5 meters/seconds. The tension of the cable is 165N - which is the magnitude of the friction force that the water and the air produces against the skier?

Since the water -skier is moving with constant velocity ( i.e. no net force), tension in the cable-frictional force=0 or tension in the cable=frictional force

Answer: magnitude ...

#### Solution Summary

Answers questions on Force, Newton's 2nd law.