A student pulls a box of books on a smooth horizontal floor with a force of 100 N in a direction of 37° above the horizontal. If the mass of the box and the books is 40.0 kg, what is the acceleration of the box?

1.9965 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S.I unit of acceleration is m/s².

From the question,

F' = ma Equation 1

Where F' = horizontal force, m = mass of the box, a = acceleration of the box.

But,

F' = FcosФEquation 2

Where F = Force applied to the box, Ф = angle of between the force and the horizontal.

Substitute equation 2 into equation 1

FcosФ = ma

make a the subject of the equation

a = FcosФ/m Equation 3

Given: F = 100 N. Ф = 37°, m = 40 kg.

Substitute into equation 3

a = 100(cos37°)/40

a = 100(0.7986)/40

a = 79.86/40

a = 1.9965 m/s²

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1