Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.
We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force
acting in the opposite direction. So Newton's second law can be rewritten as
or
since the frictional force is 15 N and we want to achieve an acceleration of
, we can substitute these values to find what is the force the man needs:
Since Astronaut and wrench system is isolated in the space and there is no external force on it
So here momentum of the system will remain conserved
so here we can say
initially both are at rest
so here plug in all values
so here the astronaut will move in opposite direction and its speed will be equal to 0.20 m/s
Explanation:
We know that that the range of the ball on the earth
therefore, range of the ball on moon
therefore,
Therefore, the range of ball will be 6 times on the moon than that on earth
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
