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Vesnalui [34]
3 years ago
11

A block with a weight of 9.0 N is at rest on a horizontal surface. A 1.2 N upward force is applied to the block by means of an a

ttached vertical string. What are the (a) magnitude and (b) direction of the force of the block on the horizontal surface
Physics
1 answer:
lesya [120]3 years ago
8 0

Answer:

a)   9 - 1.2 = 7.8 N

b) Since the force exerted by the box is it's weight, it acts in a downward direction. So the box will exert a force downward (perpendicular to the horizontal surface).

Explanation:

A 9N block would exert 9N of normal force on the horizontal plane under normal conditions. But in this case, we have a spring taking away some of the force by applying an upward force on the box.

So the force exerted by the box on the surface would now be:

a)   9 - 1.2 = 7.8 N

b) Since the force exerted by the box is it's weight, it acts in a downward direction. So the box will exert a force downward (perpendicular to the horizontal surface).

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The position of a particle moving along the x axis varies in time according to the expression x = 4t 2, where x is in meters and
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Answer:

a) X = 17.64 m

b) X = 17.64 + 4∆t^2 + 16.8∆t

c) Velocity = lim(∆t→0)⁡〖∆X/∆t〗 = 16.8 m/s

Explanation:

a) The position at t = 2.10s is:  

                 X = 4t^2

                 X = 4(2.10)^2

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b) The position at t = 2.10 + ∆t s  will be:

                 X = 4(2.10 + ∆t)^2

                 X = 17.64 + 4∆t^2 + 16.8∆t  m

c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,

                 ∆X= 4∆t^2 + 16.8∆t

Divide by ∆t on both sides:  

                ∆X/∆t =  4∆t + 16.8

 Taking the limit as ∆t approaches to zero we get:  

              Velocity =lim(∆t→0)⁡〖∆X/∆t〗 = 4(0) + 16.8

              Velocity = 16.8 m/s

 

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