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] #?

Answer 1

3.08 units

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

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Benjamin Asiedu

The work is #=1.55J#

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

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

#=1/2 + pi /3#

#=1.55J#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 3

William Audy

The work done is given by the definite integral of the force function over the given interval:

[ \text{Work} = \int_{0}^{\pi/3} (\sin(x) + 1) , dx ]

[ \text{Work} = \left[-\cos(x) + x\right]_{0}^{\pi/3} ]

[ \text{Work} = -\cos(\pi/3) + \pi/3 - (-\cos(0) + 0) ]

[ \text{Work} = \frac{1}{2} + \frac{\pi}{3} ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.