Calculating Eraser's Distance on Friction Desk

  • Thread starter Thread starter Matt1234
  • Start date Start date
  • Tags Tags
    Friction
AI Thread Summary
To calculate the distance the eraser slides on the desk, the conservation of energy principle is applied, equating the spring's potential energy to the work done against friction. The spring's energy is calculated using the formula Ee = 0.5 K x^2, where K is the spring constant and x is the compression distance. Given a spring constant of 22 N/m and a compression of 3.5 cm, the potential energy is determined. The work done against friction, represented as w = f d, uses the friction force of 0.042 N to find the distance d. By setting the spring energy equal to the friction work, the distance the eraser travels can be solved.
Matt1234
Messages
141
Reaction score
0

Homework Statement



A student uses a compressed spring of force constant 22 N/m to shoot 0.0075 kg eraser across a desk. The magnitude of the force of friction on the eraser is 0.042N. How far along the horizontal desk will the eraser slideif the spring is initially compressed 3.5 cm? Use the law of conservation of energy.

Homework Equations



Ee = .5 K x^2
Ek = 0.5 m v^2
w = f d cos ()


The Attempt at a Solution



http://img189.imageshack.us/i/lastscank.jpg/
 
Physics news on Phys.org
You don't need the kinetic energy equation for this problem. Set the spring energy equal to the friction energy and solve for the unknown d that the eraser would have to travel to make the equality true.
 
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