Question

A small object is placed at the top of an incline that is essentially frictionless. The object slides down the incline onto a rough horizontal surface, where it stops in 5.0 s after traveling 60 m. a) What is the speed of the object at the bottom of the incline and its acceleration along the horizontal surface? b) What is the height of the incline?

Answer 1

Answer

given,

time taken to stop by the object = 5 s

distance travel before stopping = 60 m

final velocity = 0

using equation of motion

v = u + at

0 = u - 5 a

$a_x = \dfrac{u}{5}$

$s = u t + \dfrac{1}{2}at^2$

$s = 5u + \dfrac{1}{2}\times \dfrac{4}{5}\times u^2$

$60 = 2.5 u$

u = 24 m/s

$a_x = \dfrac{24}{5}$

a_x = 4.8 m/s²

b) using energy conservation

$\dfrac{1}{2}mv_i^2 + \dfrac{1}{2}mv_f^2 = mg(\Delta h)$

$\dfrac{1}{2}v_i^2= g(\Delta h)$

$(\Delta h) = \dfrac{0.5 \times 24^2}{9.8}$

Δh = 29.38 m

Answer 2

(a) The speed of the object at the bottom of the incline is 24 m/s and its acceleration along the horizontal surface is 4.8 m/s².

(b) The height of the incline is 29.4 m.

Speed of the object at the bottom of the incline

The speed of the object at the bottom of the ramp is calculated as follows;

v = u + at ---(1)

where;

• u is the initial velocity = 0

v = at

v = 5a

Second equation

v² = u² + 2as

v² = 0 + 2as

v² = 2as

(5a)² = 2a(60)

25a² = 120a

25a = 120

a = 120/25

a = 4.8 m/s²

v = 5 x 4.8

v = 24 m/s

Height of the incline

The height of the incline is determined from the conservation of energy,

mgh = ¹/₂mv²

gh = ¹/₂v²

h = (v²)/(2g)

h = (24²)/(2 x 9.8)

h = 29.4 m

Learn more about conservation of mechanical energy here: https://brainly.com/question/6852965