Answer:
5.7 m
Explanation:
AD = length of the ladder = L = 8 m
AB = the position of the ladder's center of mass = (0.5) L = (0.5) 8 = 4 m
AC = distance of the climber from the bottom of the ladder = x
W = weight of the ladder = 240 N
= weight of the climber = 710 N
F = force exerted by the wall on the ladder
N = normal force acting on the ladder from the ground =?
By applying force equilibrium in the vertical direction
N = + W
N = 710 + 240
N = 950 N
μ = Coefficient of static friction = 0.55
f = static friction force on the ladder
Static friction force can be expressed as
f = μ N
f = (0.55) (950)
f = 522.5 N
The equation for force along the horizontal axis reads
F = f
F = 522.5 N
using torque equilibrium around point A
F Sin50 (AD) = W Cos50 (AB) + ( Cos50 (AC))
(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)
x = 5.7 m
Answer
Data provided:
mass of the block = 200 g = 0.2 Kg
Velocity at A = 0 m/s
Velocity at B = 8 m/s
distance of slide = 10 m
height of the block = 4 m
calculation for the block's potential energy
P = m g h
P = 0.2 x 9.8 x 4
P = 7.84 J
kinetic energy calculated as
Work done = P - KE
work = 7.84 - 6.14
work = 1.7 J
b) using the formula v² = u² + 2 a s
0 = 8² - 2 x a x 10
a = 3.2 m/s²
ma - μ mg = 0
Answer:
50.2 cm
Explanation:
We have the following data:
Height, h=3.5 m
Initial horizontal velocity,
Time, t=0.32 s
We need to determine how far the ball is from the ground after 0.32 s.
Initial vertical velocity,
Where