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# Force and Newton's 2nd law

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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 of the friction force that the water and the air produces against the skier=165 N

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?

Mass of the parachuter=m= 72 Kg
acceleration due to gravity=g= 9.8 m/s^2
Weight of the parachuter=mg= 705.6 N =72*9.8

Answer: Weight of the parachuter= 705.6 N

The ascendant force that the air exercises on the parachuter and on the parachute is equal to the weight of the parachuter since the parachuter is moving with constant velocity i.e. there is no net force therefore weight - ascendant force of the air= 0 or weight= ascendant force of the air

Answer:ascendant force of the air= 705.6 N

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?

acceleration=a= 0.175 m/s^2
mass of the car=m= 1270 Kg

Net force= ma= 222.25 N =1270*0.175

Net force=Force applied- force opposing motion
Or 222.25=4770-force opposing motion
Or force opposing motion=4770-222.25= 4547.75 N

Answer:magnitude of the force that opposes movement= 4547.75 N

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?

Initial velocity= 0 Km/hr= 0 m/s =0*1000/3600
Final velocity= 80 Km/hr= 22.2222 m/s =80*1000/3600
time required to reach the final velocity= 9.4 s

acceleration =a= rate of change of velocity = (final velocity - initial velocity)/ time taken for change in velocity
acceleration=a= 2.3641 m/s^2 =(22.2222-0)/9.4
mass of the car=m= 1060 Kg

Net force required= ma= 2506 N =1060*2.3641

Answer: force that should be exercised on the vehicle in order to obtain this acceleration= 2506 N

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?

Initial velocity= 0 m/s
Final velocity= 2.3 m/s
time required to reach the final velocity= 10.3 s

acceleration =a= rate of change of velocity = (final velocity - initial velocity)/ time taken for change in velocity
acceleration=a= 0.2233 m/s^2 =(2.3-0)/10.3
mass of the car=m= 1730 Kg

Net force required= ma= 386 N =1730*0.2233

Answer: force that should be exercised by the pulling cable= 386 N

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.

Initial velocity=u= 17.5 m/s
Final velocity=v= 0 m/s (since the car has to stop)
distance at which the car has to stop=x= 94.5 m

For constant acceleration / deceleration v^2=u^2+ 2 ax
where a is the acceleration

Or 0^2= 17.5^2+ 2* a* 94.5
or a= -17.5^2/ (2* 94.5)= -1.6204 m/s^2 (negative sign denotes deceleration)

acceleration=a= -1.6204 m/s^2
mass of the car=m= 1570 Kg

Net force required= ma= -2544 N =1570*-1.6204
(since the magnitude is negative, the force is a stopping force)

Answer: magnitude of the stopping force = 2544 N

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