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Goshia [24]
3 years ago
12

While moving a box 10.0 m across the floor, Sam pushes with a force of 150 N to the right. The force friction acting on the box

is 25.0 N to the left. What is the net work done on the box by these 2 forces
Physics
1 answer:
svet-max [94.6K]3 years ago
4 0

Answer:

Explanation:required formula is

W 1=F*S

W1=work done by Sam =?

F=force applied by sam=150N

S=displacement =10m

again

W2=F*S

W2=work done by friction =?

S=displacement =10m

F=friction =25N

W=W1-W2=net work done

please feel free to ask if you have any questions

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124.62\ \text{rad/s}

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N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}

The angular speed is 124.62\ \text{rad/s}

r = 2.98 cm

Tangential speed is given by

v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}

Tangential speed at the required point is 3.71\ \text{m/s}

Radial acceleration is given by

a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2

The radial acceleration is 1.23\ \text{km/s}^2.

t = Time = 2.06 s

Distance traveled is given by

d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}

The total distance a point on the rim moves in the required time is 20.28\ \text{m}.

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