physics problems (2)

Gravitational attraction F follows the inverse square law. That is, F is inversely proportional to the square of distance d from the earth centre .
i.e. F =k/d^2, where k is a proportional constant.

When the object is moved to a point one radius R above the earth surface, its distance from the earth centre now becomes 2R.

thus F'= k/(2R)^2 =(1/4)[k/R^2]
but [k/R^2] is just the force acting on the object when it is at earth surface.
Hence, the attractive force, which is the weight of the object, becomes 1/4 of that on earth surface.

The answer is therefore option (1), 25N
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5. Let F be the normal reaction acting on the man,
mg be the weight of the man (m being the mass and g is the acceleration due to gravity),
a be the acceleration of the lift

From Newton's second law,
net force = mass x acceleration
taking the upward direction as +ve,
we have F+(-mg) = ma
i.e. F-mg = ma

Since it is given that F-mg = 30N
thus 30 =ma
a = 30/m
because m is always +ve, thus a is +ve, this shows that the lift is accelerating upward.

The answer is option (1)