A perfect spring of spring constant 4000 N/m is used to launch a block of mass 0

Question

A perfect spring of spring constant 4000 N/m is used to launch a block of mass 0.5 kg up a 30-degree incline which has a coefficient of kinetic friction of 0.2. At its equilibrium length, the end of the spring is 1 m from the end of the ramp which is 1 m off the ground. A brick wall is 2 m away from the end of the ramp as shown in the figure. To launch the block the spring is compressed by 10 cm and released from rest. The block flies off the end of the ramp and strikes the brick wall.
(a) What is the work done by friction during the launch?

(b) What is the potential energy of the spring when compressed?

(c) What is the speed of the block as it leaves the ramp?

(d) What time is the block in the air before it hits the wall?

(e) How high above the floor does the block strike the wall?

(f) What is the kinetic energy of the block just before it hits the wall?

Explanation / Answer

A)The work done by friction will be the product the force of friction and the distance.
We already know the distance (1m) so we just need to find the magnitude of the force:

Ffric=.2FN=.2(mgcos)=.2(.5)(9.8)cos(30)

Ffric=.8487 N

W=Ffricd=.8487(1)

W=.8487 Joules

B) Enery of a spring is (1/2)kx2

PE=(1/2)(4000)(.1)2

PE=20 Joules

C)So we know the starting energy is the energy of spring, that will have to equal the sum of the other energies at the spot is leaves the ramp and the work done by friction

20=PEg+KE+W=mgh+(1/2)mv2+.8487

we can solve for v2:

v2=2[20-.8487-(.5)(9.8)(1)]/.5

v=7.55 m/s

D)To find the time we need to figure out the horizone velocity component as it leaves the 30-degree ramp and then figure out how long it will take the block to travel the horizontal distance to the wall based on this constant horizontal velocity

vx=vcos=7.55cos30

vx=6.5386 m/s

x=vxt

t=x/vx=2/6.5386

t=.3059 seconds

E)Now we just use the information we know and the following kinematic equation

y=(1/2)gt2+vsin()t+y0=-4.9(.3059)2+7.55sin(30)(.3059)+1

y=1.696 m

F) we can now say the energy of block at the edge of ramp is equal to the energy at the wall

20-.8487=KE+PEg=KE+mgh

KE=20-.8487-(.5)(9.8)(1.696)

KE=10.841 Joules

Hope that helps

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.