Question

You promised your younger brother that you would give him a ride on his sled. (Take the positive i direction to be to the right and the positive j direction to be upward.) (a) You first pull him with a force F53.21.0j) N. If the mass of the sled and boy is 59.0 kg and the coefficient of kinetic friction is 0.061, what is the acceleration of the sled? Express your answer in vector form. m/s you decite to push im from behind wih oe --21.) Determine the aceleration of th (b) After a while, you decide to push him from behind with a force F2-53.0 21.0j) N. Determine the acceleration of the sled now to see if it will be different from your answer in part (a). Express your answer in vector form m/s

Answer 1

(a) Pulling case: The expression for Newton's second law of motion is, = mã Here, F is the net force acting on the object along the inclined plane, m is the mass of the object, and a is the acceleration of the object. The weight of the system is, W = mg = (59 kg) (9.8 m/s) = 578.2 N Applying the equilibrium condition along vertical direction. N+(F)-W = 0 N+21 N -578.2 N = 0 N = 557.2 N(1) The friction acting on the object is, f = HN(-i) = (0.061)(557.2 N)(-i) = 33.9892 N(-i) The net force along horizontal is, Free = ma Ē: (i)+ f(-i)= mą, (F:- 5)(i)= ma Box-(1) = a. 53 i-33.9892 N 59 kg ą, = 0.322 m/s? ()

(b) Pushing case: Applying the equilibrium condition along vertical direction. N+F)-W = 0 N+(-21 N)-578.2 N = 0 N = 599.2 N(1) The friction acting on the object is, f = 4.N(-i) = (0.061) (599.2 N)(-i) = 36.5512 N(-1) The net force along horizontal is, Free = ma 7+()+ (-i)=ma (.:- 1)()= mą, 53 i –36.5512 N (i) = a. 59 kg a = 0.2788 m/s? (i) Hence, the conclusion drawn that for the acceleration for pushing the sled is less than the acceleration for pulling the sled.