Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
Answer:
2.4 m/s
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.08 kg)(0.5 m/s) + (0.05 kg)(0 m/s) = (0.08 kg)(-0.1 m/s) + (0.05 kg) v
0.04 kg m/s = -0.08 kg m/s + (0.05 kg) v
0.12 kg m/s = (0.05 kg) v
v = 2.4 m/s
His power output was 3 Watt (360 Joule/120 seconds). The power output can be calculated by dividing the quantity of work by the amount of second needed for the activity and also by multiplying the force amount with the velocity of the activity. The power output usually used for measuring the ability of machine for doing its job.
Answer:
6.86 N
Explanation:
Applying,
F = mg............... Equation 1
Where F = Force exerted by gravity on the mass, m = mass, g = acceleration due to gravity
Note: The Force exerted by gravity on the mass is thesame as the weight of the body.
From the question,
Given: m = 700 g = (700/1000) = 0.7 kg
Constant: g = 9.8 m/s²
Substitute these values into equation 1
F = 9.8(0.7)
F = 6.86 N
