The force applied against a moving object travelling on a linear path is given by #F(x)= sinx + 1 #. How much work would it take to move the object over #x in [ 0,pi/3] #?

2 Answers
Jan 13, 2018

Answer:

3.08 units

Explanation:

Work done W= #F.ds# = #m*(dv)/dt*ds#
= #m*dv(ds)/dt# = #m (dv)^2#
Given equation,
F = #sin x + 1#
Or, #mv (dv)/dx = sin x + 1#
Or, #mv dv = sin x dx + dx#
Integrating we get,
#m((v1)^2 - (v2)^2)= (-cos x + x)/2 #
Putting the interval we get,
#m dv ^2 = (-(1/2)+1+(pi/3))*2# i.e 3.08 units

Jan 13, 2018

Answer:

The work is #=1.55J#

Explanation:

The equation is

#"Work"="Force"xx"distance"#

Here,

The force is #F(x)=sinx+1#

Therefore,

#DeltaW=FxxDeltax#

#dW=(sinx+1)dx#

Integrating both sides

#W=int_0^(pi/3)(sinx+1)dx#

#=[-cosx+x]_0^(pi/3)#

#=(-cos(pi/3)+pi/3)-(-cos(0)+0)#

#=-1/2+pi/3+1#

#=1/2+pi/3#

#=1.55J#