More practice problems

  1. A skate boarder (60 kg) hits the bottom of a ramp going with velocity v. The ramp has the shape of a quarter-circle of radius R=3m. a) How fast does the skate boarder have to go in order to reach the top of the ramp? b) If she comes in with exactly this velocity, what is the normal force of the skate boarder on the ramp when she reaches half the maximum height?
  2. A 2.0 kg block is pushed against a spring with negligible mass and spring constant k=400 N/m, compressing it 0.22 m. When the block is released it moves along a horizontal surface and then up an incline with slope 40 degrees (ignore friction). a) What is the velocity of the block as it slides along the horizontal surface? b) How far does the block travel up the incline? c) What is the maximum acceleration of the block as it is accelerated by the spring?
  3. A ball of mass 0.2 kg is attached to a (very light) string of length 0.4 m. The ball is rotating around the vertical axis such that the string makes a constant angle of 30 degrees with the vertical. a) What is the tangential velocity of the ball? b) What is the total (kinetic plus potential energy) of the ball? What is the ratio of kinetic to potential energy? Assume that the potential energy of the ball is zero when the string is vertical.


  1. a) energy conservation: 1/2*mv^2=mgR. Get v=sqrt(2gR)= 7.67 m/sec.
    b) Energy conservation gives v=sqrt(gR). Normal force is the sum of the centripetal force and the normal component of the gravitational force: N = m*v^2/R+m*g*cos(alpha)=mg+mg/2=1.5*mg=883 N
  2. a) Energy conservation 1/2*kx^2=1/2*mv^2, so v=sqrt(k/m)x =3.11 m/sec.
    b) Energy conservation 1/2*mv^2=mgh, so h=v^2/(2*g)=0.49m. Distance traveled d=h/sin(alpha)=0.76m.
    c) Force F=kx, acceleration a=F/m=k/m*x=44 m/s^2= 4.5g
  3. a) centriptal force F_c=mv^2/r, gravitational force F_g=mg; equilibrium condition tan(alpha)=F_c/F_g, so v=sqrt(rg*tan(alpha)). Use r=l*sin(alpha), get v=sqrt(lg*sin(alpha)*tan(alpha))= sqrt(lg*0.29)=1.06 m/sec.
    b) K=1/2*m*v^2=0.11 J, U=mgh=mgl*(1-cos(alpha))=mgl*0.13=0.105 J.
    Ratio K/U=tan(alpha)*sin(alpha)/2/(1-cos(alpha))=1.07

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