Question

A projectile is fired from the ground level with an initial velocity that has a vertical component of 20 m/s and a horizontal component of 30 m/s. Find the range R. In the figure below, a constant external force P=160 N is applied to a 20.0 kg box, which is on a rough horizontal surface. While the force pushes the box a distance of 8.00 m, the speed changes from 0.500 m/s to 2.60 m/s. Find the work done by friction during this process.

Answer 1

2) V_y = 20 m/s

V_x = 30 m/s

time for which the body stays in air is given by

$t = \frac{2V_{y}}{g} = \frac{2*20}{9.8}=4.1s$

range R = V_xt = 30 x 4.1 =123 m

3) work done by force = P cos 30 (x) = 160 cos 30 x = 1108.5 joules

change in kinetic energy = (1/2) m (V_f²- V_i²) = 65.1 joules

work done by friction = work done by force - net work done = 1108.5 - 65.1=1043.4 joules

Answer 2

1) Time taken by object to reach top most point - t

In vertical direction -

v = u + at

v = 0, u = 20, g = -10

t = (v - u) / g

= 20 / 10 = 2s

Time taken by object to reach ground = 2t = 4s

In this time in horizontal direction the object covers range s = ut

u = 30, t = 4

range = ut = 30 * 4 = 120 m

2) P in horizontal direction = Pcos30 = 160(root(3) / 2) = 138.4 N

Given, v = 2.6, u = 0.5, s = 8

v² - u² = 2as

a = (2.6² - 0.5²)/(2*8)

= 0.406

F = ma = 20 * 0.406 = 8.12

Therefore, friction force = 138.4 - 8.12

= 130.28

Work done by friction = Fs = 130.28 * 8 = 1042.24J

Since force and displacement in opposite direction W = -1042.24J

Answer 3

2> We are given the vertical velocity vv=20m/s and horizontal velocity hv=30m/s.

We know, $S = ut + \frac{at^2}{2}$

For projectile to hit ground again,

S=0,u=vv,a=-9.8m/s

So, time taken, $t = \frac{2u}{-a}$

or, $t = \frac{2*20}{9.8}=4.1s$

As horizontal acceleration = 0,

using same equation, range r can be computed.

S=r, u = hv,a=0

So,range of projectile, $r = u*t = 30*4.1\ m=123\ m$

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3> Horizontal component of force, fh=$p\ cos\ 30^{\circ}$

Work done, W = Force*Distance,

or, $W = p\ cos\ 30^{\circ}\ *\ 8 = 1108.5 J$

Change in kinetic energy is given by, $K = \frac{1}{2}m(V^2_f - V^2_i)$

or, $K = 65.1\ J$

Work done by friction = total work done - work done by other forces

or, $W_{friction}=W_{total}-W_{force,p}$

or, $W_{friction}=65.1-1108.5=-1043.4\ J$ [Negative as force of friction is always in the direction opposite to the displacement]