A 3.0 kg breadbox on a frictionless incline of angle ¥è = 42 ¢ª is connected, by a cord that runs over a pulley, to a light spring of spring constant k = 110 N/m, as shown in the figure. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 9.0 cm down the incline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (if up the incline then type 1, if down the incline then type 2) of the box's acceleration at the instant the box momentarily stops?
5 answers
Angle is 42 degrees
The sum of energies is a constant.
gPE + KE + PEspring=K
initially, KE, PEspring is zero.
so gPE is k
now when it is down the plane .09m,
-.09sin42+KE+1/2 k .09^2=0
so you can figure KE.
When it stops, then KE is zero, and
x*sin42=1/2 k x^2
gPE + KE + PEspring=K
initially, KE, PEspring is zero.
so gPE is k
now when it is down the plane .09m,
-.09sin42+KE+1/2 k .09^2=0
so you can figure KE.
When it stops, then KE is zero, and
x*sin42=1/2 k x^2
Thank you, but how do you obtain the speed of the box?
K=mv^2
?
And acceleration? Sorry, I am just so confused right now
?
And acceleration? Sorry, I am just so confused right now
What do you mean by energy is a constant
0 answers