Spring and Projectile Motion Problem

AI Thread Summary
The discussion revolves around a physics problem involving a spring and projectile motion, where a block is released from a compressed spring on a frictionless tabletop. The user successfully calculated the time of flight using projectile motion principles, arriving at the equation time = sqrt((-2h)/g). They encountered difficulties determining the horizontal velocity of the block due to the lack of a spring constant but eventually realized they could use the distance traveled in relation to the time calculated. By applying equations for spring velocity and period, they were able to derive the spring constant and work done on the block. The conversation highlights the importance of connecting concepts in physics to solve complex problems effectively.
mkcompu
Messages
2
Reaction score
0
Ok, I need some help getting started with a problem involving a spring and projectile motion.

one end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance h above the floor. a block of mass m is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance x, as shown above. the block is released, follows the trajectory shown, and strikes the floor a horizontal distance d from the edge of the table. air resistance is negligible.

determine expressions for the following quantities in terms of m, x, d, h and g. notes that theses symbols do not include the sprig constant.

a) the time elapsed from the instant the block leaves the table to the instant it strikes the floor

b) the horizontal component of the velocity of the block just before it hits the floor.

c) the work done on the block by the spring.

d) the spring constant

A picture of the problem is attached.


Part A was easy, done with projectile motion. I worked it out to time = sqrt((-2h)/g).

Now I'm stuck on where to start with part B because we are not given a spring constant. I know that just the speed as it leaves the platform needs to be found because there is no acceleration on the x component.

I played around with simple harmonic motion equations but keep getting stuck on the fact that I don't know what K is or the force needed.


And for reference I'm in ap physics b.


Any help to get me started would be greatly appreciated.


Thanks.
 
Physics news on Phys.org
Hint: You figured out the time in part a. What horizontal distance did it travel in that time?
 
Doc Al said:
Hint: You figured out the time in part a. What horizontal distance did it travel in that time?


Thanks. It was so obvious, I got too caught up in the equations. I never thought I would forget that v=s/t but I just did...Thanks for the reminder.



So I got v=D/sqrt((-2h)/g)


Ok, so then It's a matter of using an equation for velocity of a spring, v=(2piX)/T and combining it with T=2pisqrt(m/k), I can find K [got part D done right there!] and then using w= 1/2kx^2.


Sounds good to me.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top