A conveyer belt is moving at a constant speed of 2 ms^(-1) A box is gently dropped on it. The coefficient of friction between them is mu = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g= 10 ms^(-2), is

A conveyer belt is moving at a constant speed of 2 ms^(-1) A box is ge

1.	A conveyer belt is moving at a constant speed of 2 ms^(-1) A box is gently dropped on it. The coefficient of friction between them is mu = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g= 10 ms^(-2), is
Answer» <html><body><p>`0.4 m`<br/>1.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> m<br/>0.6m<br/>zero</p>Solution :Forceof Friction` f^(-)mu <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a> ` <br/> ` thereforea = (f)/( m )= (mu mg)/( m)= mu g= 0.5xx 10= <a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> <a href="https://interviewquestions.tuteehub.com/tag/ms-549331" style="font-weight:bold;" target="_blank" title="Click to know more about MS">MS</a>^(-2)`<br/> <a href="https://interviewquestions.tuteehub.com/tag/using-7379753" style="font-weight:bold;" target="_blank" title="Click to know more about USING">USING</a>`v^2- u^2= 2aS `<br/> ` 0^2 -2^2 =2 (-5) xx SimpliesS= 0.4m`</body></html>