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An object with a mass of #7 kg# is hanging from a spring with a constant of #2 (kg)/s^2#. If the spring is stretched by # 8 m#, what is the net force on the object?

Answer 1

William Audy

#P.E=1/2 kx^2#

#=1/2*64*2#

#=64# Joule
TOTAL FORCE ACTING #=mass*g#
#=70N#

WORK DONE =#F*S#
THEREFORE #64=F*8#
#F=8N#

NET FORCE#=70-8=62N#

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Answer 2

Adrian Ayala

The net force acting on the object can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.

[ F = -kx ]

Where:

( F ) is the force exerted by the spring,
( k ) is the spring constant, and
( x ) is the displacement from the equilibrium position.

Given:

( k = 2 , (\text{kg/s}^2) )
( x = 8 , \text{m} )

Substitute the given values into the formula:

[ F = -(2 , \text{kg/s}^2)(8 , \text{m}) = -16 , \text{N} ]

Since the force is in the opposite direction of the displacement, the net force on the object is ( -16 , \text{N} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.