M g H sin37 = (1/2)*kX^2
X = 0.20 m
k = 200 N/m
M = 1.0 kg
g = 9.8 m/s^2
Solve for H. This neglects additional work done during spring compression, but should be close enough.
The exact equation is
M g (H+X) sin37 = (1/2)*kX^2
A 1.0-kg block is released from rest at the top of a frictionless incline that makes and angle of 37 degrees with the horizontal. An unknown distance down the incline from the point of release, there is a spring with k=200 N/m. It is observed that the mass is brought momentarily to rest after compressing the spring .20m. What distance does the mass slide from the point of release until it is brought momentarily to rest?
5 answers
Use simple harmonic motion
W(g)=U(s)
mgdsin(37)=1/2kx^2
d=kx^2/2mgsin(37)
d=0.68m
mgdsin(37)=1/2kx^2
d=kx^2/2mgsin(37)
d=0.68m
Use the Lay of conservation of mechanics:
W + U + K = U + K + E(lost)
The body started from rest so work done on it is 0, it has potential energy, but we started with 0 initial velocitY. After the motion, we come to rest so the final velocity is 0, and its frictionless so no energy is lost.
we get:
U = U(spring)
mgh = 0.5*k*x^2
To get the h, make a right angle triangle corresponding to the initial position of the spring and we know: SOH CAH TOA so dsin(37)=h
(1)g(dsin(37)) = 0.5(200)(0.2)^2
All you need to do is make d the subject and you get 0.68m.
W + U + K = U + K + E(lost)
The body started from rest so work done on it is 0, it has potential energy, but we started with 0 initial velocitY. After the motion, we come to rest so the final velocity is 0, and its frictionless so no energy is lost.
we get:
U = U(spring)
mgh = 0.5*k*x^2
To get the h, make a right angle triangle corresponding to the initial position of the spring and we know: SOH CAH TOA so dsin(37)=h
(1)g(dsin(37)) = 0.5(200)(0.2)^2
All you need to do is make d the subject and you get 0.68m.
thamks smart human