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AleksAgata [21]
3 years ago
7

If a force does not act parallel to the resulting displacement, what is the effect on the work done by the force? 

Physics
1 answer:
m_a_m_a [10]3 years ago
8 0
Then only the component of the force that's parallel to the displacement
is used to calculate the work.  The component of force that's perpendicular
to the displacement  doesn't move through any distance at all, so its contribution
to the total work is zero.
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The semi truck travels at an initial speed of 69.545 meters per second downwards.

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In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v (1)

Where:

m_{S}, m_{C} - Masses of the semi truck and the car, measured in kilograms.

v_{S}, v_{C} - Initial velocities of the semi truck and the car, measured in meters per second.

v - Final speed of the system after collision, measured in meters per second.

If we know that m_{S} = 2200\,kg, m_{C} = 2000\,kg, v_{C} = 45\,\frac{m}{s} and v = -15\,\frac{m}{s}, then the initial velocity of the semi truck is:

m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}

v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}

v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}

v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})

v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s}  \right)

v_{S} = -69.545\,\frac{m}{s}  

The semi truck travels at an initial speed of 69.545 meters per second downwards.

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