A 20 Newton force is applied to a 2kg block that rests on a desk. The block is on the verge of sliding. What is the coefficient of static friction?
Since the block is on the verge of sliding, we know frictional force, Fr, is equal to the applied force, F. We also know that in this case Fr is equal to the coefficient of static friction, mus, times the normal force, N or Fr=F=mus*N. In this case the normal force is the weight of the block. On earth gravity is 9.8 m/s^2 so the normal force N=2kg(9.8m/s^2)=19.6N. Putting it together, 20N=mus*19.6N. Solving for mus, mus=1.02
Evaluate the following, integral(sin(x)*x dx)
The best method for solving this problem is through integration by parts. Let u=x and dv=sin(x). Therefore, du=dx v=-cos(x). Integration by parts takes the form, u*v-integral(v*du). This leads to -x*cos(x)-integral(-cos(x)dx). The integral of -cos(x) is -sin(x). Thus, the answer is -x*cos(x)+sin(x)
If f(x)=x^2-5x+4, what is f(3)-f(2)?
f(3)=3^2-5*3+4 f(3)=9-15+4 f(3)=-2 f(2)=2^2-5*2+4 f(2)=4-10+4 f(2)=-2 f(3)-f(2)=-2-(-2) f(3)-f(2)=0