# Calculating with 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 parachute 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. What is the force?

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##### Solution Summary

The solution involves various calculations using Newton's 2nd Law, including calculations for friction force, weight, gravitational force and acceleration.

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1.) Because, the box has zero acceleration, therefore, along the plane (see attached figure),

Force of the cord F = mgsin(27) + f

(Because, F-mgsin(27)-f = m*a, a =0)

friction force,

f = F - mgsin(27) = 365 - 55.2*9.8*sin(27) = 119.41 N --Answer

2.) Because, the speed is constant, therefore,

the friction force = Applied force = 165 N --Answer

(Because, F-f = m*a, a =0)

3.) mass of man M = 72 kg

v(constant) = 9.1 m/s

mass of parachute m = 6.5 kg

a.) Parachuer's weight,

W = Mg ...

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- BEng, Allahabad University, India
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