A 5-kg block being pushed across a table by a force P has an acceleration of 4 m

Question

A 5-kg block being pushed across a table by a force P has an acceleration of 4 m/s. a) What is the net force acting on the block? b) IfP 15 N, what is the magnitude of the frictional force acting upon the block? 9. 10. A passenger of mass 100 kg is riding in an elevator while standing on a platform scale as shown in Figure 1. The scale only reads the normal force N it exerts on the passenger. (Note that 1 lb 0.453 kg 4.45 N). Using g- 9.8 m/s2 a) What does the scale read in pounds (lb) when the elevator is not moving? b) What does the scale read in pounds (lb) when the elevator is moving up with an acceleration of 3.0 m/s2? What does the scale read in pounds (lb) when the elevator is moving down with an acceleration of 3.0 m/s?? What does the scale read in pounds (lb) when the elevator is moving down with constant velocity? c) d) Figun

Explanation / Answer

(9)

a)

Net force, F = ma = 5 x 4 = 20 N

b)

Friction(f) is opposite to the applied force P

So, F = P - f = ma

=> f = P - ma = 15 - 20 = -5 N

=> |f| = 5 N

(10)

Treating upwards as +ve and downwards as -ve.

Normal force = N

Weight = W = mg

If the elevator moves with an acceleration 'a', Newtons second law gives

N - W = ma = 100a, where a = acceleration.

(a)

a = 0

=> N = W = mg = 100 x 9.8 = 980 N = 980/4.45 lbs = 220.225 lbs

(b)

a = 3 m/s2

=> N = W + ma = 100 x 9.8 + 100 x 3 = 1280 N = 1280/4.45 lbs = 287.640 lbs

(c)

a = -3 m/s2

=> N = W + ma = 100 x 9.8 - 100 x 3 = 680 N = 680/4.45 lbs = 152.809 lbs

(d)

Contant velocity means zero acceleration. Thus

a = 0

=> N = W + ma = 100 x 9.8 + 0 = 980 N = 220.225 lbs

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