The work done by the force is 140 J.
Calculation:
It is given that, an 80-N crate is pushed a distance of 5.0 m upward along a smooth incline that makes an angle of
with the horizontal. The force pushing the crate is parallel to the slope and the acceleration is 1.5
.
It is required to find the work done by the force.
The free-body diagram is shown below,
Here, W=80 N, a=-1.5 ![\text{ m/s}^2](https://tex.z-dn.net/?f=%5Ctext%7B%20m%2Fs%7D%5E2)
We can write the net force equation as,
![\begin{gathered}F-W \sin 30^{\circ}=m a \\F-m g \sin 30^{\circ}=\left(\frac{W}{g}\right) a\end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7DF-W%20%5Csin%2030%5E%7B%5Ccirc%7D%3Dm%20a%20%5C%5CF-m%20g%20%5Csin%2030%5E%7B%5Ccirc%7D%3D%5Cleft%28%5Cfrac%7BW%7D%7Bg%7D%5Cright%29%20a%5Cend%7Bgathered%7D)
Here W is the weight of the object, m is the mass of the body, and F is the applied force.
Thus,
![\begin{aligned}F-(80 \mathrm{~N}) \sin 30^{\circ} &=\left(\frac{80 \mathrm{~N}}{9.8 \mathrm{~m} / \mathrm{s}^{2}}\right)\left(-1.5 \mathrm{~m} / \mathrm{s}^{2}\right) \\F &=27.755 \mathrm{~N}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DF-%2880%20%5Cmathrm%7B~N%7D%29%20%5Csin%2030%5E%7B%5Ccirc%7D%20%26%3D%5Cleft%28%5Cfrac%7B80%20%5Cmathrm%7B~N%7D%7D%7B9.8%20%5Cmathrm%7B~m%7D%20%2F%20%5Cmathrm%7Bs%7D%5E%7B2%7D%7D%5Cright%29%5Cleft%28-1.5%20%5Cmathrm%7B~m%7D%20%2F%20%5Cmathrm%7Bs%7D%5E%7B2%7D%5Cright%29%20%5C%5CF%20%26%3D27.755%20%5Cmathrm%7B~N%7D%5Cend%7Baligned%7D)
It is known that the work done by the force is calculated as the multiplication of the applied force and the displacement.
Thus the work one is,
![\begin{aligned}W &=(27.755 \mathrm{~N})(5 \mathrm{~m}) \\& \approx 140 \mathrm{~J}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DW%20%26%3D%2827.755%20%5Cmathrm%7B~N%7D%29%285%20%5Cmathrm%7B~m%7D%29%20%5C%5C%26%20%5Capprox%20140%20%5Cmathrm%7B~J%7D%5Cend%7Baligned%7D)
Thus the last option is the correct one.
Learn more about work done by the force here,
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