Physics 11: How to Calculate Spring Compression with an Applied Force of 6.0 N

  • Thread starter Thread starter roxxyroxx
  • Start date Start date
  • Tags Tags
    Physics Physics 11
AI Thread Summary
To calculate the spring compression when a 2.8 kg box moving at 1.9 m/s hits a spring, the initial kinetic energy (KE) of the box must be equated to the potential energy (PE) stored in the spring. The spring constant (k) is determined to be 50 N/m based on a 6.0 N force compressing the spring by 12 cm. Applying the energy conservation principle, the equation 0.5mv² = 0.5kx² leads to the correct compression distance of 0.45 m. The key is to use the box's initial kinetic energy before contact with the spring. This approach ensures accurate calculations of spring compression.
roxxyroxx
Messages
48
Reaction score
0

Homework Statement



A spring can be compressed 12 cm by an applied force of 6.0 N. A 2.8 kg box sliding along a frictionless surface approaches this spring at a speed of 1.9 m/s. How far will the spring be compressed when the box comes to a stop?

Homework Equations



Ep = 0.5kx2

0.5mv2 = 0.5kx2

The Attempt at a Solution



I worked out that k = 50 N/m
0.5mv2 = 0.5kx2
0= 25x2

but this doesn't work out because x should be 0.45 m. help?
 
Physics news on Phys.org
When the box comes to a stop all of the Kinetic Energy will be converted to the potential energy of the spring.

Use the formula you had only put the initial KE of the box in before it touches the spring.

And all will be well.
 
great thanks !
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
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