Question

A concrete slab of mass 200kg pulled 10m up a slop at an angle of 30 degree to the horizontal ,coefficient of friction (kinetic =0.6) find the work done

Answer 1

Answer:

= 26.94 m/s

Explanation:

given,

angle of inclination = 30°

mass of the sleigh = 200 kg

coefficient of kinetic friction = 0.2

height of inclination = 10 m

pull force be = 5000 N

now,.

T - f_s - mg sin \theta = m aT−f

s

−mgsinθ=ma

T - \mu N - mg sin \theta = m aT−μN−mgsinθ=ma

T - \mu mg - mg sin \theta = m aT−μmg−mgsinθ=ma

a = \dfrac{T}{m} - \mu g - g sin \thetaa=

m

T

−μg−gsinθ

a = \dfrac{5000}{200} - \0.2\times 9.8 - 9.8 \times sin 30^0a=

200

5000

−\0.2×9.8−9.8×sin30

0

a = 18.14\ m/s^2a=18.14 m/s

2

L = \dfrac{10}{sin 30}L=

sin30

10

L = 20 m

v² = u² + 2 as

v² = 0 + 2 x 18.14 x 20

v = 26.94 m/s