Question
4.10 please
A 68.5-kg skater moving initially at 2,40 m/s rough horizontal ice comes to rest uniformly in 3.52 s due to friction from the ice. What force does friction exert on the skater? You walk into an elevator, step onto a scale, and push the "up" button. You also recall that your normal weight is 625 N. Start answering each of the following questions by drawing a free diagram. (a) If the elevator has an acceleration of magnitude 250 m/s^2, what does the scale read? (b) If you start holding a 385-kg package by a light vertical string, what will be the tension in this string once the elevator begins accelerating? A box rests on a frozen pond, which serves as a frictionless horizontal surface. If a fisherman applies a horizontal force with 4.18 magnitude 48.0 N to the box and produces an acceleration of magnitude 3.00 m/s^2, what is the mass of the box? A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. (a) What is the mass of the block of ice? (b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s? A hockey puck with mass 0.160 kg is at rest at the origin of (x = 0) on the horizontal, frictionless surface of the rink. At time t = 0 a player applies a force of 0.250 N to the puck, parallel to the x-axis: he continues to apply this force until t = 2.00 s. (a) What are the position and speed of the puck at t = 2.00 s? (b) What are position and speed of the puck at t = 5.00 s, what are the position on and speed of the puck at t = 7.00 s? A crate fate with mass 32.5 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 140 N. (a) What acceleration is produced? (b) How far does the crate travel in 10.0 s? (c) What is its speed at the end of 10.0 s? A 4.50-kg toy cart undergoes an acceleration in a straight line (the x-axis).The graph in Fig. E4.13 shows this acceleration as a function of time
Explanation / Answer
4.10: Applying d = v0 t + a t^2 / 2
11 = 0 + a(5^2)/2
a = 0.88 m/s^2
from newton's 2nd law,
F = m a
m = 80 / 0.88 = 90.9 kg .......Ans(a)
(b) speed after 5 sec,
v = v0 + a t = 0 + (0.88 x 5) = 4.4 m/s
now it will move with constant velocity.
distance travelled = vt = 4.4 x 5 = 22 m ......Ans
