What is the force constant k of a spring which is stretched
a) 2 mm by a force of 4 N,
b) 4 cm by a mass of 200 g?
Answers
Answer:
To find:
Force constant of a spring in the following cases:
2 mm by a force of 4 N
4 cm by a mass of 200 g
Calculation:
We will consider that at the given extension , the spring force is equal and opposite to the applied force.
In the 1st case:
\rm\therefore \: force = - kx∴force=−kx
\rm = > \: - 4= - k(2 \times {10}^{ - 3} )=>−4=−k(2×10
−3
)
\rm = > \: 4= k(2 \times {10}^{ - 3} )=>4=k(2×10
−3
)
\boxed{ \rm = > \: k = 2000 \: N {m}^{ - 1} }
=>k=2000Nm
−1
In the 2nd case:
\rm\therefore \: force = - kx∴force=−kx
\rm = > \: - \dfrac{200}{1000} \times g = - k(4 \times {10}^{ - 2} )=>−
1000
200
×g=−k(4×10
−2
)
\rm = > \: \dfrac{200}{1000} \times 10 = k(4 \times {10}^{ - 2} )=>
1000
200
×10=k(4×10
−2
)
\rm = > \: \dfrac{2000}{1000} = k(4 \times {10}^{ - 2} )=>
1000
2000
=k(4×10
−2
)
\rm = > \: 2 = k(4 \times {10}^{ - 2} )=>2=k(4×10
−2
)
\rm = > \: k = 0.5 \times {10}^{2}=>k=0.5×10
2
\boxed{ \rm = > \: k =50 \: N {m}^{ - 1} }
=>k=50Nm
−1
Hope It Helps.