Question

A rigid body of mass 0.3 kg is taken slowly up an inclined plane of
length 10 m and height 5 m (assuming the applied force to be parallel to
the inclined plane), and then allowed to slide down to the bottom again.
The co-efficient of friction between the body and the plane is 0.15.
Using g = 9.8 m/s2 find the
(a) work done by the gravitational force over the round trip.
(b) work done by the applied force over the upward journey
(c) work done by frictional force over the round trip.
(d) kinetic energy of the body at the end of the trip?

PLZ TELL WHAT F APPLIED UPWARD EXPLAIN PLZZZ ​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{W.D_{g}=0\:J}}}$

$\green{\tt{\therefore{W.D_{up}=29.4(0.5+0.075\sqrt{3})\:J}}}$

$\green{\tt{\therefore{W.D_{fr}=-4.51\sqrt{3}\:J}}}$

$\green{\tt{\therefore{K.E=14.7\:J}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies Length \: of \: inclined = 10 \: m \\ \\ \tt: \implies Height \: of \: wedge = 5 \: m \\ \\ \tt: \implies Mass \: of \:block (m) = 0.3 \: kg \\ \\ \tt: \implies Coefficient \: of \: friction( \mu) = 0.15 \: \mu \\ \\ \red{\underline{\bold{To \: find :}}} \\ \tt: \implies Work \: done \: by \: gravitational \: force(W.D_{g})=? \\ \\ \tt: \implies Work \: done \: by \: upward \: applied\: force(W.D_{up}) = ? \\ \\ \tt: \implies Work \: done \: by \: frictional \: force(W.D_{fr}) = ? \\ \\ \tt: \implies Final \: kinetic \: energy \: of \: block \:(K.E) =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies sin \: \theta = \frac{p}{h} \\ \\ \tt: \implies sin \: \theta = \frac{5}{10} \\ \\ \tt: \implies sin \: \theta = \frac{1}{2} \\ \\ \tt: \implies \theta =30 \degree \\ \\ \green{ \star} \tt \: Component \: of \: normal = mg \: cos \: \theta \: and \: mg \: sin \: \theta \\ \\ \green{ \star} \tt \: Frictionl\: force = \mu N = \mu \: mg \: cos \: \theta \\ \\ \green{ \star} \tt \:Downward \: force = mg \: sin \: \theta \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies W.D_{g} =2\times f \: s \: cos \theta \\ \\ \tt: \implies W.D_{g} =mg \: sin \: \theta \times (10-10) \times cos \: 0 \degree \\ \\ \tt: \implies W.D_{g} =0.3\times 9.8\times sin\:\theta\times 0\times 1$

$\green{ \tt: \implies W.D_{g} =0 \: J} \\ \\ \bold{Again : } \\ \green{\star}\tt\:f=mg\:sin\:\theta + \mu\:N\\\\\tt: \implies W.D_{up} =f \: s\: cos \: \theta \\ \\ \tt: \implies W.D_{up} =mg \:sin \theta+\mu\:mg\:cos\:theta \times 10 \times cos \: 0 \degree \\ \\ \tt: \implies W.D_{up} =0.3 \times 9.8 \times \frac{1}{2}\times 0.15\times0.3\times 9.8\times \frac{ \sqrt{3} }{2} \times 10 \times 1\\ \\ \green{\tt: \implies W.D_{up} = 29.4(0.5+0.075\sqrt{3}) \: J} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies W.D_{fr} =2f \: s \:cos \theta \\ \\ \tt: \implies W.D_{fr} = 2\times \mu N \: \times 10 \times cos \: 180 \degree \\ \\ \tt: \implies W.D_{fr} =2\times 0.15 \times 0.3 \times 9.8 \times \frac{ \sqrt{3} }{2} \times 10 \times - 1 \\ \\ \green{\tt: \implies W.D_{fr} = - 4.51 \sqrt{3} \: J}$

$\green{ \star } \tt \: u = 0 \: m/s \\ \\ \green{ \star } \tt \:a = g \: sin \: \theta\:m/s^{2} \\ \\ \green{ \star } \tt \: h = 10 \: m \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {v}^{2} = {0 }^{2} + 2 \times 9.8 \times \frac{1}{2} \times 10 \\ \\ \tt: \implies {v}^{2} = 98 \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies K.E = \frac{1}{2} m {v}^{2} \\ \\ \tt: \implies K.E= \frac{1}{2} \times 0.3 \times 98 \\ \\ \green{\tt: \implies K.E = 14.7 \: J}$

Answer 2

(a) As the displacement over the round trip is zero so, work done by the gravitational force over the round trip is zero.

(b) work done by the applied force over the upward journey

Wup=F×d

=(mgsinθ+f)h

=mg(sinθ+μcosθ)h

=0.3×9.8×10((105)+0.15(1075))

=29.4(0.5+0.129)

=18.51J

(c) work done by frictional force over the round trip

W=2fh

=2(μmghcosθ)

=2×0.15×0.3×9.8×1075×10

=7.638J

(d) kinetic energy of the body at the end of the trip

K=21mv2

From newton’s second law

mgsinθ−f=ma

mgsinθ−μmgcosθ=ma

a=g(sinθ−μcosθ)

a=9.8(0.5−0.15×1075)

a=3.62ms