A 2.0 kg box is pulled along a frictionless, horizontal surface by a force that makes an angle of 30 degrees with the ground. If the box is pulled a distance of 4.0 meters and
changes speed from 1.0 m/s to 3 m/s, what was the magnitude of the force on the box?

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Ответ:

henryisasum7846

05.04.2020

Physics

13.8 N

$41.44^{\circ}$

Explanation:

$F_1=6\ \text{N}$

$F_2=8\ \text{N}$

$F_1\cos\theta_1\hat{i}+F_1\sin\theta_1\hat{j}\\ =6\cos30^{\circ}+6\sin30^{\circ}\hat{j}\\ =5.2\hat{i}+3\hat{j}$

$F_2\cos\theta_2\hat{i}+F_2\sin\theta_2\hat{j}\\ =8\cos50^{\circ}+8\sin50^{\circ}\hat{j}\\ =5.14\hat{i}+6.13\hat{j}$

$F_R=F_1+F_2=10.34\hat{i}+9.13\hat{j}$

$|F_R|=\sqrt{10.34^2+9.13^2}=13.8\ \text{N}$

The magnitude of the resultant is 13.8 N

Direction is given by

$\tan^{-1}=\dfrac{y}{x}=\tan^{-1}\dfrac{9.13}{10.34}=41.44^{\circ}$

The angle of the resultant is $41.44^{\circ}$

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A 2.0 kg box is pulled along a frictionless, horizontal surface by a force that makes an angle of 30 degrees with the ground. If the box is pulled a distance of 4.0 meters and changes speed from 1.0 m/s to 3 m/s, what was the magnitude of the force on the box?