integral of F(x) dx
= a x^2/2 + b x^3/3
at 3 this is
9a/2 + 27b/3
at 1 it is
a/2 + b/3
subtract
8a/2 - 26b/3
now put in a and b
40/2 + 52/3
etc
An object , constrained to move along the x-axis is acted upon by a force F(x)=ax + bx^2 where a=5N/m , b=-2m/s
The object is observed directly from x=1m to x=3.0 m... How much work is done by the force?
4 answers
Final answer
Work=2.67J
Work=2.67J
F(x)= dU/dx, du= F(x)dx
=> dU= (ax + bx²)dx. Integrating we get
U= ax²/2 + bx³/3. At x = 1m we have
U= a/2 + b/3.....(1). At x = 3m we have
U= 9a/2 + 27b/3.....(2). Subtracting we hv
U= 8a/2 + 26b/3. Solving we get
6U= 24a + 52b. Substituting the values of a & b, we have
6U= 120 - 104 => 6U =16, U = 2.67 joules
=> dU= (ax + bx²)dx. Integrating we get
U= ax²/2 + bx³/3. At x = 1m we have
U= a/2 + b/3.....(1). At x = 3m we have
U= 9a/2 + 27b/3.....(2). Subtracting we hv
U= 8a/2 + 26b/3. Solving we get
6U= 24a + 52b. Substituting the values of a & b, we have
6U= 120 - 104 => 6U =16, U = 2.67 joules
2.67j