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leva [86]
2 years ago
15

Which force below does the most work? All three displacements are the same. The 10 N force. The 8 N force The 6 N force. They al

l do the same work. Sin60º = 0.87 cos60º = 0.50
Physics
1 answer:
Daniel [21]2 years ago
5 0

THe question is missing parts. Here is the complete question.

Which force below does the most work? All three displacements are the same.

A. The 10 N force

B. The 8 N force

C. The 6 N force

D. They all do the same work

note: sin60° = 0.87

        cos60° = 0.5

Answer: C. The 6 N force

Explanation: <u>Work</u> <u>(</u>τ<u>)</u> is the transfer of energy to or from a system by moving an object.

Work is dependent of the force we applied to the object, the displacement it creates and the angle between the force and the displacement. In other words:

\tau = F.d.cos \theta

and its unit is joule [J].

For the 10 N force, angle is 90°. Cosine of 90° is 0. Therefore:

\tau = 10.d.cos 90

\tau = 0

For the 8 N force, angle is 60°. Then:

\tau = 8.d.cos 60

\tau = 8d(0.5)

\tau = 4d

For the 6 N force, angle is 0, because vectors displacement and force are pointing to the same direction:

\tau = 6.d.cos 0

\tau=6d(1)

\tau= 6d

Comparing the work done, the force that does the most work is force 6N.

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At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is
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Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

\dfrac{1}{2}mv^2=mgl

v=\sqrt{2gl}

Put the value into the formula

u=\sqrt{2\times9.8\times50.0\times10^{-2}}

u=3.13\ m/s

The initial speed of the ball u_{1}=3.13\ m/s

The initial speed of the block u_{2}=0

(a). We need to calculate the speed of the ball after collision

Using formula of collision

v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13

v_{1}=-2.01\ m/s

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0

v_{2}=1.11\ m/s

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

8 0
3 years ago
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