A short 10-g string is used to pull a 50-g toy across a frictionless horizontal surface. If a

acceleration=force/total mass=.03N/.06kg= .5m/s^2

forceonthetoy=mass*a=.05*.5=.025N

bob is right, the answer is C

a is constant on both thus finding a will answer the question

acceleration=force/total mass=.03N/.06kg= .5m/s^2

forceonthetoy=mass*a=.05*.5=.025N

Well, let's calculate the force of the string on the toy. We know that the force applied to the free end of the string is 3.0 x 10^(–2) N.

Since there is no friction on the surface, the force of the string pulling the toy should be equal to the force applied.

So, the force of the string on the toy would be 3.0 x 10^(–2) N.

Therefore, the correct answer would be D) 3.0 x 10^(–2) N.

To solve this question, we need to apply Newton's second law, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration. In this case, the force applied by the string on the toy is given by the equation:

Force = mass × acceleration

The mass of the toy is 50 g, which is equivalent to 0.050 kg. The force applied by the string is 3.0 × 10^(-2) N, and the acceleration is given by dividing the force by the mass:

acceleration = Force / mass

Now we can substitute the values into the equation:

acceleration = 3.0 × 10^(-2) N / 0.050 kg

Calculating this gives:

acceleration = 0.60 m/s^2

Since the surface is frictionless, there are no other forces acting on the toy apart from the force applied by the string. Therefore, the force of the string on the toy is equal to the mass of the toy multiplied by the acceleration:

Force = mass × acceleration

Force = 0.050 kg × 0.60 m/s^2

Evaluating this expression gives:

Force = 0.030 N

Therefore, the force of the string on the toy is 0.030 N, which is equivalent to 3.0 × 10^(-2) N.

So the correct answer is option D) 3.0 × 10^(-2) N.