Physics, 02.07.2019 13:30 diana4393

A19.7 kg block is dragged over a rough, horizontal surface by a constant force of 188 n acting at an angle of 28.6 ◦ above the horizontal. the block is displaced 48.7 m, and the coefficient of kinetic friction is 0.103.find the work done by the 188 n force. the acceleration of gravity is 9.8 m/s 2 . answer in units of j find the magnitude of the work done by the force of friction. answer in units of j. a) what is the change in kinetic energy of the crate? answer in units of j. b) what is the speed of the crate after it is pulled the 9.28 m? answer in units of m/s.

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This question involves the concepts of work, the law of conservation of energy, frictional force, normal reaction, and kinetic energy.

a) Work done by the force is "8038.46 J".

b) Work done by the frictional force is "517.96 J".

c) Change in kinetic energy is "7250.5 J"

d) Speed of the crate after covering 9.28 m distance is "27.63 m/s".

The work done by the applied force (188 N) is given as follows:

W = Fd Cosθ

where,

W = Work done = ?

F = Applied force = 188 N

d = distance moved = 48.7 m

θ = Angle between applied force and horizontal surface = 28.6°

Therefore,

W = (188 N)(48.7 m) Cos 28.6°

W = 8038.46 J

Writing equilibrium equation in the vertical direction:

$W = N + FSin\theta$

where,

W = weight of the block = mg = (19.7 kg)(9.81 m/s²) = 193.26 N

N = Normal Reaction Force = ?

F = Applied force = 188 N

Therefore,

$193.26\ N = N+(188\ N)Sin\ 28.6^o$

N = 103.26 Newtons

Now, frictional force can be given as follows:

$f =\mu N$

where,

f = frictional force = ?

μ = coefficient of friction = 0.103

Therefore,

$f=(0.103)(103.26\ N)$

f = 10.64 N

Thus the work done by this frictional force can be calculated as follows:

$W_f=fd\\W_f=(10.64\ N)(48.7\ m)\\W_f=517.96\ J$

From the law of conservation of energy:

$Change\ in\ Kinetic\ Energy = Net\ Work\\\Delta K.E=W-W_f = 8038.46\ J-517.96\ J\\$

ΔK.E = 7520.5 J

The final speed of the crate can be given by the following formula:

$\Delta K.E = 7520.5 J\\\\\frac{1}{2}m(v_f^2-v_i^2)=7520.5\ J\\\\$

where,

m = mass = 19.7 kg

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

$\frac{1}{2}(19.7\ kg)(v_f^2-(0\ m/s)^2) = 7520.5\ J\\\\v_f^2=\frac{(2)(7520.5\ J)}{19.7\ kg}\\\\v_f=\sqrt{763.5 m^2/s^2}\\\\v_f = 27.63\ m/s$

Learn more about the law of conservation of energy here:

link

The attached picture explains the law of conservation of energy.

Answer from: johnb9163

Part a)

W = 8038.5 J

Part b)

W_f = -516.6 J

Part c)

$\Delta K = 7521.9 J$

Part d)

v_f = 27.6 m/s

Explanation:

As per force equation on the block we have

$mg = F_n + F sin\theta$

$19.7 \times 9.8 = F_n + 188 sin28.6$

F_n = 103 N

Now we have

$F_x - F_f = F_{net}$

$188 cos28.6 - \mu F_n = ma$

165 - (0.103)(103) = 19.7 a

a = 7.84 m/s^2

Part a)

Now we have to find work done by applied force

W = F.d

$W = F d cos\theta$

$W = 188 \times 48.7 cos28.6$

W = 8038.5 J

Part b)

Now work done by friction force is given as

$W_f = -\mu F_n .d$

W_f = -(0.103)(103)(48.7)

W_f = -516.6 J

Part c)

By work energy theorem we have

Total work done = change in kinetic energy

$8038.5 - 516.6 = W_{net}$

$W_{net} = 7521.9 J$

$\Delta K = 7521.9 J$

Part d)

Final speed of the block is given as

$\frac{1}{2}mv_f^2 = 7521.9$

$\frac{1}{2}(19.7)v_f^2 = 7521.9$

v_f = 27.6 m/s

A19.7 kg block is dragged over a rough, horizontal surface by a constant force of 188 n acting at an

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