Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Hello! Assuming that the only force acting on the mass is 30N...
Fnet = 30N
Fnet = ma (mass x acceleration)
ma = 30N
a = 30N / m
a = 30N / 7kg
a = 4.2857 m/s^2
a = 4 m/s^2
I hope this helps!
work is force x distance = 25 x 0.4
= 2.5x4 = 10joules
pwer would be 10j/2s watts .... 5 watts
Answer:
Explanation:
From the question we are told that:
Initial Speed 
Time 
Angle 
Generally the Newton's equation for motion is mathematically given by
consider the motion along the X-direction
X = horizontal displacement = 80 m
= initial velocity along the x-direction = v Cos60
t = time of travel
using the equation
X = t
80 = (v Cos60) (t)
t = 160/v eq-1
consider the motion in vertical direction :
Y = vertical displacement = 20 m
= initial velocity in Y-direction = v Sin60
a = acceleration = - 9.8 m/s²
t = time of travel = 160/v
using the equation
Y = t + (0.5) a t²
20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²
v = 32.5 m/s
