
You pull a short refrigerator with a constant force across a greased (frictionless) floor, either horizontally (case 1) or tilted upward at an angle (case 2).
(a)What is the ratio of the refrigerators speed in case 2 to its speed in case 1 if you pull for a certain time t?
(b) What is this ratio if you pull for a certain distance d?
Answer
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Hint: In this question, we have given two cases in which the refrigerator is pulled along a horizontal plane and tilted upwards. We can find the given ratio by using Newton’s laws of motion and the concept of acceleration, velocity and displacement in each case. Also, the resolution of force vectors.
Complete step by step answer:
Consider the refrigerator moving along +X direction and starting from rest. The force acting on the refrigerator is the X-component of the force when it is tilted upwards through an angle $\theta $.
(a) We have to find the ratio of refrigerator in given two cases i.e. case 1- refrigerator is pulled with a constant force across horizontal floor and case 2 - refrigerator is pulled with a constant force tilted upwards at angle $\theta $.Using Newton’s 2nd law, we have
${F_x} = ma = m\left( {\dfrac{v}{t}} \right)$
\[\therefore v = \left( {\dfrac{{F\cos \theta }}{m}} \right)t\]
Denoting ${v_1}$ - speed in 1st case , ${v_2}$ - speed in 2nd case
And \[{\theta _1}\] - angle in 1st case , ${\theta _2}$ - angle in 2nd case
Given, \[{\theta _1} = 0,{\theta _2} = \theta \]
Taking the ratio of ${v_1}$ and \[{v_2}\] , we get
$\therefore \dfrac{{{v_2}}}{{{v_1}}} = \cos \theta $
(b) Now, we have to find the distance when the refrigerator is pulled for a certain distance $d$.Using Newton’s 2nd law and displacement, we get
${F_x} = m\left( {\dfrac{{{v^2}}}{{2\Delta x}}} \right)$
$\therefore v = \sqrt {2\left( {\dfrac{{F\cos \theta }}{m}} \right)} $
Taking the ratio of ${v_1}$ and \[{v_2}\] , we get
$\therefore \dfrac{{{v_2}}}{{{v_1}}} = \sqrt {\cos \theta } $
Note: It must be noted that there should be separate equations for time and displacement in both cases. The angle between the refrigerator and the floor is zero because it is pulled without any inclination with respect to the floor. The force exerted during both cases must be taken non-zero and constant, in order to cancel it for the ease while formulating the equation. Trigonometric ratios must be used in order to split the components of force.
Complete step by step answer:
Consider the refrigerator moving along +X direction and starting from rest. The force acting on the refrigerator is the X-component of the force when it is tilted upwards through an angle $\theta $.
(a) We have to find the ratio of refrigerator in given two cases i.e. case 1- refrigerator is pulled with a constant force across horizontal floor and case 2 - refrigerator is pulled with a constant force tilted upwards at angle $\theta $.Using Newton’s 2nd law, we have
${F_x} = ma = m\left( {\dfrac{v}{t}} \right)$
\[\therefore v = \left( {\dfrac{{F\cos \theta }}{m}} \right)t\]
Denoting ${v_1}$ - speed in 1st case , ${v_2}$ - speed in 2nd case
And \[{\theta _1}\] - angle in 1st case , ${\theta _2}$ - angle in 2nd case
Given, \[{\theta _1} = 0,{\theta _2} = \theta \]
Taking the ratio of ${v_1}$ and \[{v_2}\] , we get
$\therefore \dfrac{{{v_2}}}{{{v_1}}} = \cos \theta $
(b) Now, we have to find the distance when the refrigerator is pulled for a certain distance $d$.Using Newton’s 2nd law and displacement, we get
${F_x} = m\left( {\dfrac{{{v^2}}}{{2\Delta x}}} \right)$
$\therefore v = \sqrt {2\left( {\dfrac{{F\cos \theta }}{m}} \right)} $
Taking the ratio of ${v_1}$ and \[{v_2}\] , we get
$\therefore \dfrac{{{v_2}}}{{{v_1}}} = \sqrt {\cos \theta } $
Note: It must be noted that there should be separate equations for time and displacement in both cases. The angle between the refrigerator and the floor is zero because it is pulled without any inclination with respect to the floor. The force exerted during both cases must be taken non-zero and constant, in order to cancel it for the ease while formulating the equation. Trigonometric ratios must be used in order to split the components of force.
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