hows a 170 g hamster sitting on an 830 g wedge-shaped block. The block, in turn, rests on a spring scale. An extra-fine lubricating oil having μs=μk=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.

The complete question with the scale is missing, so i have attached it.

930 g

Explanation:

The block surface is frictionless, therefore, the only force that can possibly be applied to the block from the hamster will be the normal force.

Now, the force of gravity on the hamster acts downward and equals mg. Hence, The normal component will be;

F_n = mgcosθ.

Now, if we Assume that the scale does not respond to horizontal force component, the downward component of the normal force will be expressed as;

F_d = F_n(cosθ)

The scale measures the downward component.

Earlier, we saw that; F_n = mgcosθ

Thus; F_d = mgcosθ × cosθ

F_d = mg(cos²θ)

Now, when the friction was high, the measurement of the scale was;

(m+ M)g

Where ;

m = mass of hamster

M = mass of block

Therefore, when the friction is removed the scale measurement is;

mgcos²θ + Mg or (mcos²θ + M)g

But this is in units of Newton's.

Thus;

Measurement in kg is;

(mcos²θ + M)

We are given;

m = 170 g = 0.17 kg

M = 830 g = 0.83 kg

From the attached scale image,

θ = 40°

Thus;

Measurement = (mcos²θ + M) = ((0.17 cos²40) + 0.83) = 0.93 kg or 930g

found in the nucleus

has mass of one amu

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