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MissTica
3 years ago
14

. An 8-ton truck is initially going up a 20° ramp with a speed of 90 km/hr. The truck, then, brakes reaching a speed of 36 km/hr

after 3 seconds. Assuming constant deceleration determine (a) the magnitude of the braking force, (b) the additional time required for the truck to stop. Neglect air resistance and rolling resistance.
Physics
1 answer:
marshall27 [118]3 years ago
6 0

Answer:

a) F = 13166.468\,N, b) t = 2\,s.

Explanation:

a) The equations of equilibrium for the truck are: (x' is the axis parallel to ramp, y' is the axis perpendicular to ramp)

During braking:

\Sigma F_{x'} = - m\cdot g \cdot \sin \theta - F = m\cdot a

\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0

The deceleration experimented by the truck is:

a=\frac{(36\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} )-(90\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} )}{3\,s}

a = -5\,\frac{m}{s^{2}}

The braking force is:

F= - m\cdot (g\cdot \sin \theta+a)

F = -(8000\,kg)\cdot [(9.807\,\frac{m}{s^{2}} )\cdot \sin 20^{\textdegree}-5\,\frac{m}{s^{2}} ]

F = 13166.468\,N

b) The additional time required for the truck to stop is:

t = \frac{0\,\frac{m}{s}-(36\,\frac{km}{h} )\cdot(\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} ) }{-5\,\frac{m}{s^{2}} }

t = 2\,s.

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