The force acting in the front direction is the 130N.
The frictional force is acting backwards 30N.
1) The net force is 130N - 30N = 100N
2) s = ut + (1/2)at^2 u = 0, Start from rest, s = 25m t =5.
25 = 0*5 + (1/2)* a * 5^2.
25 = 0 + 25/2 * a.
25 = (25/2)a. Divide 25 from both sides.
1 = (1/2)* a. Cross multiply.
2 = a.
a = 2 m/s^2.
3) Mass of the box
Net Force, F = ma
100 = m*2. Divide both sides by 2.
100/2 = m
50 = m.
m = 50 kg.
4) Final velocity , v = u + at.
v = 0 + 2*5 = 10 m/s.
Kinetic Energy, K = (1/2) * mv^2.
= 1/2 * 50 * 10 * 10.
= 2500 J.
Answer:
Explanation:
Given data
Radius R=6.80 m
Velocity v=18 km/h =5 m/s
First we need to set up for Force in vertical (y) and horizontal (x)
∑Fy=0=TCosα-W
∑Fx=ma=Fc=TSinα
Solve Vertical force equation for T,substituting mg for W
Substitute expression for Fc and T into horizontal force and simplify it we get
