How long must a tow truck apply a force of 600 N to increase the speed of a 1,500 kg car at rest to 2 m/s?
1 answer:
The time the truck must apply the given force to increase its speed to given value is 5 s.
The given parameters;
- <em>applied force, F = 600 N</em>
- <em>mass of the truck, m = 1,500 kg</em>
- <em>speed of the truck, v = 2 m/s</em>
The force applied to the truck is determined by Newton's second law of motion; <em>which states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.</em>
F = ma
Thus, the time the truck must apply the given force to increase its speed to given value is 5 s.
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<u>Given data:</u>
m= 50 Kg,
W= m×g = 50 × 9.81 = 490.5 N
ramp angle (α) = 30 degrees,
coefficient of friction (μs) = 0.5773,
Push at an angle (Θ) = 40 degrees,
Determine: Push to get box move up (P)=?
From the figure,
Resolving the forces along the plane
W sinα + μs.R = P cos Θ --------------------- (i)
Resolving the forces perpendicular to the inclined plane
W cosα = R+Psin Θ => R= W cosα - Psin Θ -------------- (ii)
Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>
<em>Pmin = W sin( α +Θ )</em>
<em> = W[ sin α.Cos Θ + cos α.sin Θ]</em>
<em> = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>
<em> = 460.9 N</em>
<em>Minimum push required to move the box up the ramp is 460.9 N</em>
