PLEASE WRITE THIS IN COMPUTER, NOT BY HAND. Solve: Show the procedure clear, com

Question

PLEASE WRITE THIS IN COMPUTER, NOT BY HAND.

Solve:

Show the procedure clear, complete and organized. Please do not skip any step or procedure, put everything. Do not use formulas, equations and advanced varials, this is the kind of introduction to physics, that everything is very basic.

1. A constant force of 40.0 N is applied to a 6.00 kg box by moving it on a horizontal surface, free of friction, at a distance of 9.00 meters. (a) Calculate the work done by force in the box. (b) If the force is applied to the box at an angle of 30 degrees, in the same distance, what is the work that forces the box?

2. A 750 kg car increases its speed from 40.0 mph to 60.0 mph. Find the change in the kinetic energy of the car.

3. A climber of 56.6 kg part of an elevation of 1270 m and climbs to the top of a peak of 2500 m in height. Determine the change in the mountaineer's potential energy.

4. A car accelerates from rest until it reaches a speed of 15 m / s in 8.0 seconds. If its mass is 1000 kg, determine the power in the motor.

Explanation / Answer

1. (a) Given: Force (F) = 40 N, Distance (d) = 9 m, Mass (m) = 6 Kg

Since the motion of the box is linear, distance = displacement = 9 m

We know that Work (W) = Force (F) x Displacement (d)

                                                = 40 x 9 = 360 J                                 

(b) Here, force acts at a certain angle. We shall resolve the Force vector horizontally, which gives

FHorizontal = F Cos (30) = 40 x Cos (30) = 34.64 N

Work = FHorizontal x Displacement

                   = 34.64 x 9 = 311.76 N

2. Given: Mass (m) = 750 Kg,

Initial Speed (u) = 40 mph = 40 x 1.61 Km/hr = 40 x 1.61 x 5/18 m/s = 17.89 m/s

Final Speed (v) = 60 mph = 60 x 1.61 Km/hr = 60 x 1.61 x 5/18 m/s = 26.83 m/s

Then, Initial Kinetic energy = ½ mu2

Final Kinetic Energy = ½ mv2

Hence, Change in Kinetic Energy = ½ m(v2 – u2) = ½ x 750 x (26.832 – 17.892) = 149924 J = 149.9 KJ

3. Given : Mass (m) = 56.6 Kg, Initial Height (Hi) = 1270 m, Final Height (Hf) = 2500 m

Change in potential energy = mg (Hf - Hi)

= 56.6 x 9.8 x (2500 – 1270)

= 682256.4 J = 682.26 KJ

4. Given: Mass (m) = 1000 Kg, Initial speed (u) = 0, Final Speed (v) = 15 m/s, t = 8 s

We know that Work (W) = Change in Kinetic energy

                                        = ½ m(v2 – u2) = ½ x 1000 x (225 – 0) = 112500 J

Now Power is Work done per unit time

Hence, Power = W/t = 112500/8 = 14062.5 W = 14.06 KW

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