Question

$FIGURECP7.57$shows a$200g$hamster sitting on an $800g$wedge-shaped block. The block, in turn, rests on a spring scale. An extra-fine lubricating oil having ${\mu }_s={\mu }_k=0$is sprayed on the top surface of the block, causing the hamster to slide down. Friction between the block and the scale is large enough that the block does not slip on the scale. What does the scale read, in grams, as the hamster slides down?

Short answer

The hamster slides down in scale is about$920g$

Step-by-step solution

Given information

mass of hamster $m=200g$

mass of the block $M=800g$

Angle of wedge block$\theta ={40}^0$

Explanation

The hamster accelerates down the block with $a=g\cos \theta$

The vertical component of this acceleration is: $a_y=g{\cos }^2\left(\theta \right)$

So the downward force exerted by the hamster on the block and thus onto the scale is

$F_h=m{a}_y$

The net force exerted by the hamster on the block and thus onto the scale is

$$\begin{gathered} F=Mg+mg{\cos }^2\left(\theta \right) \\ \Rightarrow F=0.8\times 9.8+0.2\times 9.8\times {\cos }^2\left(40\right) \\ \Rightarrow F=9.0N \end{gathered}$$

Therefore, the net force in gram can be calculated as

$$\begin{gathered} F=m_{s}g=9.0N \\ {m}_s=\frac{9}{9.8}\times 1000=920g \end{gathered}$$

Therefore, the hamster slides down in scale about$920g$