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otez555 [7]
2 years ago
15

On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max

pulls on the sled with 12 n of force, directed at an angle of 15° above the ground. when he comes to a recently plowed section of road, he continues to pull the sled with the same force across the road while the road exerts a frictional force of 4 n on sled. what is the net work done on the sled while max pulls it 5 m across the road?

Physics
1 answer:
sesenic [268]2 years ago
7 0

Work done by a given force is given by

W = F.d

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = W_1 = Fdcos\theta

W_1 = 12*5cos15

W_1 = 57.96 J

Now similarly work done by frictional force

W_2 = Fdcos\theta

W_2 = 4*5cos180

W_2= -20 J

Now total work done on sled

W_{net}= W_1 + W2

W_{net} = 57.96 - 20 = 37.96 J

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A woman living in a third-story apartment is moving out. Rather than carrying everything down the stairs, she decides to pack he
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T = 480.2N

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8 0
2 years ago
A 0.300kg glider is moving to the right on a frictionless, ­horizontal air track with a speed of 0.800m/s when it makes a head-o
e-lub [12.9K]

Answer:

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

0.010935 J

0.0858675 J

Explanation:

m_1 = Mass of first glider = 0.3 kg

m_2 = Mass of second glider = 0.15 kg

u_1 = Initial Velocity of first glider = 0.8 m/s

u_2 = Initial Velocity of second glider = 0 m/s

v_1 = Final Velocity of first glider

v_2 = Final Velocity of second glider

As momentum and Energy is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.3-0.15}{0.3+0.15}\times 0.8+\frac{2\times 0.15}{0.3+0.15}\times 0\\\Rightarrow v_1=0.27\ m/s

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.3}{0.3+0.15}\times 0.8+\frac{0.3-0.15}{0.3+0.15}\times 0\\\Rightarrow v_2=1.067\ m/s

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

Kinetic energy is given by

K=\frac{1}{2}m_1v_1^2\\\Rightarrow K=\frac{1}{2}0.3\times 0.27^2\\\Rightarrow K=0.010935\ J

Final kinetic energy of first glider is 0.010935 J

K=\frac{1}{2}m_2v_2^2\\\Rightarrow K=\frac{1}{2}0.15\times 1.07^2\\\Rightarrow K=0.0858675\ J

Final kinetic energy of second glider is 0.0858675 J

6 0
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