1. A spring-loaded gun is fixed with its barrel pointing at 450 to the horizontal direction. Its barrel is 1.0 m long and when relaxed, the spring is the same length. The spring constant is 1000.0 N/m. A 1.0 g projectile is fired with the spring initially compressed by 0.5 m. While traveling inside the barrel, the edges apply a constant frictional force of 2.0 N to the projectile. You may neglect air resistance.(A) What is the speed of the projectile as it leaves the barrel of the gun?(B). How high above the end of the barrel does the projectile reach before beginning to fall?We take the bottom of the gun to be at height 0. Call the mass M.Initially, the height of the projectile is (0.5 m) cos 450 .gravitional PE= Mg (0.5 m) cos 450Elastic PE = kx2 = (1000 N/m)(0.5 m)2KE=0When it leaves the barrel of the gun, the height of the projectile is (1.0 m) cos 450gravitional PE= Mg (1.0 m) cos 450Elastic PE = 0KE= Mv2.Work done by friction = - (2.0 N)(0.5 m)Conservation of energyChange in KE + Change in PE = work done( Mv2) + (Mg (1.0 m) cos 450 - Mg (0.5 m) cos 450 - (1000 N/m)(0.5 m)2) = - (2.0 N)(0.5 m)v2 = (124 J - Mg (0.5 m) cos 450 )(2/M)Version 1: M= 0.001 kg, v= 498 m/sVersion 2: M= 0.002 kg, v= 352 m/sVersion 3: M= 0.004 kg, v= 249 m/sVersion 4: M= 0.005 kg, v= 223 m/sFor the height, we have vy,i = v sin 45Final vy,f = 0v2y,i - v2y,f = 2 g hh = (v2 sin2450 )/(2g)Version 1: M= 0.001 kg, v= 498 m/s, h=6326 mVersion 2: M= 0.002 kg, v= 352 m/s, h=3163 mVersion 3: M= 0.004 kg, v= 249 m/s, h= 1581Version 4: M= 0.005 kg, v= 223 m/s, h=1265 m2. The coefficient of kinetic friction between any two surfaces is k. The pulley is massless. If M is moving downward, find the acceleration.For the mass M, we find Mg-T = MaFor the mass m, the net force perpendicular to the plane is N-mg cos and this should be zero since the mass moves on the plane.So N=mg cos .The frictional force is f= k N = k mg cos .He net force along the plane is T-mg sin - k mg cos = maAdd the equations to cancel T, and solve for aa= (Mg -mg sin - k mg cos )/(m+M)

3. A block of mass m is pressed against a wall by a force F acting upward at an angle to the vertical as shown. The coefficient of static friction is s. Find the minimum force F that will prevent the block from sliding downward.There is a normal force from the wall to the right, and a frictional force acting upward (the block is about to slide downward).Net force in horizontal direction = N- F sin and this is zero since the block is static. So N= F sin .Maximum frictional force f = s N = s F sin Net force in vertical direction = F cos + f mgSet to zero because block is staticF cos + s F sin mg = 0F = mg/( cos + s sin )