A cart loaded with bricks has a total mass
of 13.4 kg and is pulled at constant speed by
a rope. The rope is inclined at 21.1° above
the horizontal and the cart moves 17.2 m on
a horizontal floor. The coefficient of kinetic
friction between ground and cart is 0.3.
The acceleration of gravity is 9.8 m/s2.
What is the normal force exerted on the
cart by the floor?
Answer in units of N.
2.0 m Since we've been told to ignore friction, then we can also ignore gravity since all it's doing is contributing to the normal force which is only of interest if we're paying attention to friction. So how fast will 13 N of force accelerate 34 kg of mass? Since a newton is kg*m/s^2 and we want to get m/s^2, a simple division should be able to cancel out the kg. So: 13 kg*m/s^2 / 34 kg = 0.382352941 m/s^2 The distance an object moves under constant acceleration is d = 0.5 AT^2 So let's substitute the known values and calculate the distance. d = 0.5 AT^2 d = 0.5 0.382352941 m/s^2 * (3.2 s)^2 d = 0.191176471 m/s^2 * 10.24 s^2 d = 1.957647059 m Rounding to 2 significant figures gives 2.0 meters
mass of cart with bricks
coefficient of kinetic friction
it is given that cart is moving with constant velocity so net force on it is Zero
if T is the tension in rope
Net normal Reaction