- #1
Buffu
- 849
- 146
Homework Statement
Question :- A block of mass ##5 kg## is attached to a spring. The spring is stretched by ##10 cm## under the load of the block. A impulse is provided to the block such that it moves up with a velocity of ##2 m/s##. Find the height it will rise.
Homework Equations
##F_s = kx##
##x = 0.1m##
##v = 2m/s##
##m = 5kg##
##x_2## - height raised from the mean position of the spring.
The Attempt at a Solution
We can get spring constant by equating spring force and weight of the object.
$$kx = mg $$
$$k = {5 \times 10 \over 1/10} = 500 \hspace{20 mm} (1)$$
Now initial total energy of the system :-
$$TE_i = {m v^2 + kx^2\over 2}$$
Now using given data and from ##(1)##.
$$TE_i = {5 \times 4 + 500\times 1/100\over 2} = {25 \over 2}$$
final total energy :-
Velocity at maximum will be zero.
$$TE_f = {mg(x + x_2) + kx_2^2\over 2} \hspace{20 mm} (2)$$
Using conservation of energy ,##(1)## and ##(2)## we get,
$${25 \over 2} = {mg(x + x_2) + kx_2^2\over 2}$$
$$20x_2^2 + 2x_2 -0.8 = 0$$
##x_2 = 0.15##
thus total height is ##0.25 m##.
The answer is ##0.20 m##. I think the answer is incorrect, since this is pretty straight forward question.