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torisob [31]
3 years ago
6

A student pulls horizontally on a block with a spring scale. The block reads 24.5 Newtons before the block starts to move. It re

ads 23.7 Newtons as she pulls the block at a constant velocity. How should she explain this?
Physics
1 answer:
Ahat [919]3 years ago
7 0

Answer:

The coefficient of friction causes the force on the object to be less than its initial reading on the spring scale.

Explanation:

Since the block reads 24.5 N before the block starts to move, this is its weight. Now, when the block starts to move at a constant velocity, it experiences a frictional force which is equal to the force with which the student pulls.

Now, since the velocity is constant so, there is no acceleration and thus, the net force is zero.

Let F = force applied and f = frictional force = μN = μW where μ = coefficient of friction and N = normal force. The normal force also equals the weight of the object W.

Now, since F - f = ma and a = 0 where a = acceleration and m = mass of block,

F - f = m(0) = 0

F - f = 0

F = f

Since the force applied equals the frictional force, we have that

F =  μW and F = 23.7 N and W = 24.5 N

So, 23.7 N = μ(24.5 N)

μ = 23.7 N/24.5 N

μ = 0.97

Since μ = 0.97 < 1, the coefficient of friction causes the force on the object to be less than its initial reading on the spring scale.

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To pull a 57 kg crate across a horizontal frictionless floor, a worker applies a force of 230 N, directed 34° above the horizont
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a.  W_w=235\ J\\b. W_g=-343.54\ J\\c. F_N=463.100\ N\\d.  W_t=235\ J

Explanation:

Given: that,

Angle of inclination of the surface, \theta=34^{\circ}

mass of the crate, m=57\ kg

Force applied along the surface, F=230\ N

distance the crate moves after the application of force, s=1.1\ m

a) work done = F× s

 work done = 230 × 1.1

 work done = 253 J

b) Work done by the gravitational force:

W_g=m.g\times h

where:

g = acceleration due to gravity

h = the vertically downward displacement

Now, we find the height:

h=s\times sin\ \thetah=1.1\times sin\ 34^{\circ}h=0.615\ m

So, the work done by the gravity:

W_g=57\times 9.8\times (-0.615) \\= - 343.54 J

∵direction of force and displacement are opposite.

= - 343.54J

c)

The normal reaction force on the crate by the inclined surface:

F_N=m.g.cos\ \thetaF_N=57\times 9.8\times cos\ 34F_N=463.100\ N

d)

Total work done on crate is with respect to the worker:

W_t=235\ J

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