An object of mass 6 kg. is resting on a horizontal surface. A horizontal force
of 15 N is constantly applied on the object. The object moves a distance of
100m in 10 seconds.
(a) How much work does the applied force do?
(b) What is the kinetic energy of the block after 10 seconds?
(c) What is the magnitude and direction of the frictional force (if there is
any)?
(d) How much energy is lost during motion?
note from questioner: there has to be friction as the distance covered on a frictionless surface would be 125 m [ x - x⁰ = v⁰t + (1/2)*at²] but here the distance is 100 m ... but I am not sure about this so help would be appreciated
Answers
Explanation:
(a). W = F * s
= 15*100
= 1500 joule
(b). KE = 1/2 mv²
= 1/2 * 6 * (d/t)
= 1/2* 6* (100/10)
= 1/2* 6* 10
= 30
(c). frictional force acts in the opposite direction of applied force and acc. to Newton's third law of motion every action has an equal and opposite direction . So force applied by friction is equal to the aplied force but due to large mass of box it is difficult to observe ( I'm quite unsure about answer of this part)
(d). ( I don't know the answer of this part so kindly ignore it and search at Google )
...
Answer:
1) 1500 J (2) 300 J (3) μ.60 N (coefficient of friction is required to solve this completely) (4) 300 J
Explanation:
1) work donw = W = f.d = 15 x 100 = 1500 J
2) kinetic energy = K.E = 1/2 mv^2
since, value of velocity is not given, first we'll calculate it by the formula v = d/t
v = 100/10 = 10m/s
K.E = 1/2 x 6 x 10^2
K,E = 1/2 x 6 x 100
K.E = 1/2 x 600
K.E = 300 J
3) Frictional force = F = μ. mg
F= μ. 6 x 10
F = μ. 60
where "μ" is the coefficient of friction and depends on the nature of surface.
* Direction of frictional force is always opposite to the direction of motion.
4) loss in potential energy (P.E) = gain in kinetic energy (K.E)
loss in potential energy = 300 J