
A package of mass $56kg$ is pushed up a ramp with a horizontal force of $450N$. The length of the ramp is $3.00m$ and the angle between the ramp and the horizontal is ${36^\circ }$. If the final speed of the package is $5.0m{s^{ - 1}}$. Calculate the initial velocity of the package.
A) $6.7m{s^{ - 1}}$
B) $8.5m{s^{ - 1}}$
C) $9.7m{s^{ - 1}}$
D) $4.5m{s^{ - 1}}$
Answer
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Hint: In order to find the solution of the given question, first of all we need to find the various forces that are acting on the body. After that we need to balance the equation and write the equation for the same. From the equation formed we can find the value of acceleration and then finally using Newton’s equation of motion, we can find the required value of the initial velocity.
Complete step by step solution:
First of all let us find the forces that are acting on the body.
Since the angle of inclination is given and the force is acting in horizontal direction, so the horizontal component of force will be acting in upward direction.
So, we can write,$F\cos {36^\circ }$in upward direction.
Also, the force due to the weight of the body will be acting in downward direction to balance the force acting in upward direction that can be written as,$mg\sin {36^\circ }$.
Now to balance all the forces we can write,
$\Rightarrow F\cos {36^\circ } - mg\sin {36^\circ } = ma$……………..(i)
The mass of the body is given as,$m = 56kg$
And the force acting on the body is given as,$F = 450N$
Now, we need to put the values in the equation (i). So, after putting the values in equation (i), we get,
$\Rightarrow 450\cos 36 - 56(9.81)\sin 36 = 56a$
$ \Rightarrow 364.05 - 323 = 56a$
$ \Rightarrow 41.05 = 56a$
$\therefore a = 0.73m{s^{ - 1}}$
Now, we need to use Newton’s third equation of motion to find the value of the initial velocity.
$\Rightarrow {v^2} - {u^2} = 2as$
$ \Rightarrow {5^2} - {u^2} = 2 \times 0.73 \times 3$
$\Rightarrow - {u^2} = 4.38 - 25 = - 20.62$
$\therefore u = \sqrt {20.62} = 4.54m{s^{ - 1}}$
Therefore, the required value of the initial velocity is $4.54m{s^{ - 1}}$.
Hence, option (D), i.e. $4.54m{s^{ - 1}}$ is the correct choice for the given question.
Note: For any vector quantity we have its two components i.e. vertical component and the horizontal component. For the vertical component we take sin of the angle made with the vector and for the horizontal component we take cosine of the angle made with the vector.
Complete step by step solution:
First of all let us find the forces that are acting on the body.
Since the angle of inclination is given and the force is acting in horizontal direction, so the horizontal component of force will be acting in upward direction.
So, we can write,$F\cos {36^\circ }$in upward direction.
Also, the force due to the weight of the body will be acting in downward direction to balance the force acting in upward direction that can be written as,$mg\sin {36^\circ }$.
Now to balance all the forces we can write,
$\Rightarrow F\cos {36^\circ } - mg\sin {36^\circ } = ma$……………..(i)
The mass of the body is given as,$m = 56kg$
And the force acting on the body is given as,$F = 450N$
Now, we need to put the values in the equation (i). So, after putting the values in equation (i), we get,
$\Rightarrow 450\cos 36 - 56(9.81)\sin 36 = 56a$
$ \Rightarrow 364.05 - 323 = 56a$
$ \Rightarrow 41.05 = 56a$
$\therefore a = 0.73m{s^{ - 1}}$
Now, we need to use Newton’s third equation of motion to find the value of the initial velocity.
$\Rightarrow {v^2} - {u^2} = 2as$
$ \Rightarrow {5^2} - {u^2} = 2 \times 0.73 \times 3$
$\Rightarrow - {u^2} = 4.38 - 25 = - 20.62$
$\therefore u = \sqrt {20.62} = 4.54m{s^{ - 1}}$
Therefore, the required value of the initial velocity is $4.54m{s^{ - 1}}$.
Hence, option (D), i.e. $4.54m{s^{ - 1}}$ is the correct choice for the given question.
Note: For any vector quantity we have its two components i.e. vertical component and the horizontal component. For the vertical component we take sin of the angle made with the vector and for the horizontal component we take cosine of the angle made with the vector.
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