nalock befipulled across a table by a horizontal force of 100 N, experiences a a
ID: 2040664 • Letter: N
Question
nalock befipulled across a table by a horizontal force of 100 N, experiences a a) What is the net force acting on the block? b) What is the acceleration? 10. A passenger of mass 95 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). Use g 98 m/s? a) What does the scale read in pounds (lb) when the elevator is not moving? b) What does the scale read in pounds (b) when the elevator is moving up with an acceleration of 2.5 m/s?? What does the scale read in pounds (lb) when the elevator is moving down with an acceleration of 2.5 m/s? What does the scale read in pounds (lb) when the elevator is moving down with constant velocity? e) d)Explanation / Answer
(9)
a) Let the direction of the horizontal pull force be +ve and its opposite direction be -ve. Thus
Fa = Horizontal pull force = 100 N
f = Frictional Force = -10 N. Sign is -ve because f is opposite to Fa
(a)
Net force, F = Fa + f = 100 - 10 = 90 N.
(b)
According to newton's second law,
F = ma
=> a = F/m = 90/6 = 15 m/s2
(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 = 95a, where a = acceleration.
(a)
a = 0
=> N = W = mg = 95 x 9.8 = 931 N = 931/4.45 lbs = 209.213 lb
(b)
a = 2.5 m/s2
=> N = W + ma = 95 x 9.8 + 95 x 2.5 = 1168.5 N = 1168.5/4.45 lbs = 262.584 lb
(c)
a = -2.5 m/s2
=> N = W + ma = 95 x 9.8 - 95 x 2.5 = 693.5 N = 693.5/4.45 lbs = 155.843 lb
(d)
Contant velocity means zero acceleration. Thus
a = 0
=> N = W + ma = 95 x 9.8 + 0 = 931 N = 931/4.45 lbs = 209.213 lb
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