A body of mass 2 kg travels according to the law $x (t) = p t + q t^{2} + r t^{3}$ where $p = 3 m s^{- 1}, q = 4 m s^{- 2}$ and $r = 5 m s^{- 3}$ . The acting on the body at t=2 seconds is

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Answer 1

Here, $x (t) = p t + q t^{2} + r t^{3}$
Where $p = 3 m s^{- 1}, q = 4 m s^{- 2}, r = 5 m s^{- 3}$ and m=2 kg

Velocity, $v = \frac{d x}{d t} = \frac{d}{d t} (p t + q t^{2} + r t^{3}) = p + 2 q t + 3 r t^{2}$

Acceleration, $a = \frac{d v}{d t} = 2 q + 6 r t$

At $t = 2 s, a = 2 (4 m s^{- 2}) + 6 (5 m s^{- 2}) (2 s)$
= $8 m s^{- 2} + 60 m s^{- 2} = 68 m s^{- 2}$

The force acting on the body of mass 2 kg is
$F = m a = (2 k g) (68 m s^{- 2}) = 136 N$

Answer 2

Here,

x (t) = p t + q t^{2} + r t^{3}

Where

p = 3 m s^{- 1}, q = 4 m s^{- 2}, r = 5 m s^{- 3}

and m=2 kg

Velocity,

v = \frac{d x}{d t} = \frac{d}{d t} (p t + q t^{2} + r t^{3}) = p + 2 q t + 3 r t^{2}

Acceleration,

a = \frac{d v}{d t} = 2 q + 6 r t

At

t = 2 s, a = 2 (4 m s^{- 2}) + 6 (5 m s^{- 2}) (2 s)

=

8 m s^{- 2} + 60 m s^{- 2} = 68 m s^{- 2}

The force acting on the body of mass 2 kg is

F = m a = (2 k g) (68 m s^{- 2}) = 136 N

A body of mass 2 kg travels according to the law $x (t) = p t + q t^{2} + r t^{3}$ where $p = 3 m s^{- 1}, q = 4 m s^{- 2}$ and $r = 5 m s^{- 3}$ . The acting on the body at t=2 seconds is

136 N

134 N

158 N

68 N

A body of mass $5$ kg is acted upon by a variable force. The force varies with the distance covered by the body. What is the speed of the body when the body has covered $25$ m? Assume that the body starts from rest.

A $60.0 k g$ wagon experiences a variable net force in the forward direction according to the graph pictured below. If it starts from rest, what is its speed after $9$ seconds?

Same force 'F' acts on two bodies of different masses 9 kg and 25 kg which are initial at rest. then the ratio of time required by the two objects to acquire equal final velocity is.

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Power of the only force acting on a particle of mass $m = 1$ kg moving in straight line depends on its velocity as $P = v^{2}$ where v is m/s and P is in watt. If initial velocity of the particle is $1 m / s,$ then the displacement of the particle in $2$ second will be:

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A body of mass 2 kg travels according to the law x(t)=pt+qt2+rt3 where p=3ms−1,q=4ms−2 and r=5ms−3. The acting on the body at t=2 seconds is

136 N

134 N

158 N

68 N

Here, x(t)=pt+qt2+rt3Where p=3ms−1,q=4ms−2,r=5ms−3 and m=2 kg Velocity, v=dtdx​=dtd​(pt+qt2+rt3)=p+2qt+3rt2 Acceleration, a=dtdv​=2q+6rt At t=2s,a=2(4ms−2)+6(5ms−2)(2s)=8ms−2+60ms−2=68ms−2 The force acting on the body of mass 2 kg is F=ma=(2kg)(68ms−2)=136N

A body of mass 5kg is acted upon by a variable force. The force varies with the distance covered by the body. What is the speed of the body when the body has covered 25m? Assume that the body starts from rest.

A 60.0kg wagon experiences a variable net force in the forward direction according to the graph pictured below. If it starts from rest, what is its speed after 9 seconds?

Same force 'F' acts on two bodies of different masses 9 kg and 25 kg which are initial at rest. then the ratio of time required by the two objects to acquire equal final velocity is.

The acceleration produced in a body by a force depends only upon its

Power of the only force acting on a particle of mass m=1kg moving in straight line depends on its velocity as P=v2 where v is m/s and P is in watt. If initial velocity of the particle is 1m/s, then the displacement of the particle in 2 second will be:

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A body of mass 2 kg travels according to the law $x (t) = p t + q t^{2} + r t^{3}$ where $p = 3 m s^{- 1}, q = 4 m s^{- 2}$ and $r = 5 m s^{- 3}$ . The acting on the body at t=2 seconds is

A body of mass $5$ kg is acted upon by a variable force. The force varies with the distance covered by the body. What is the speed of the body when the body has covered $25$ m? Assume that the body starts from rest.

A $60.0 k g$ wagon experiences a variable net force in the forward direction according to the graph pictured below. If it starts from rest, what is its speed after $9$ seconds?

Power of the only force acting on a particle of mass $m = 1$ kg moving in straight line depends on its velocity as $P = v^{2}$ where v is m/s and P is in watt. If initial velocity of the particle is $1 m / s,$ then the displacement of the particle in $2$ second will be: